User fernando muro - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:33:02Z http://mathoverflow.net/feeds/user/12166 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/133675/a-beck-chevalley-type-condition/133701#133701 Answer by Fernando Muro for A Beck-Chevalley type condition Fernando Muro 2013-06-13T22:14:54Z 2013-06-13T22:14:54Z <p>The answer is <em>always</em>, provided you choose compatible 2-cells filling the first two diagrams. Probably, you didn't see it because you removed from notation these 2-cells. Let us give a name to the natural isomorphism for the first square:</p> <p>$$\zeta\colon k^\ast f^\ast\cong h^\ast g^\ast$$</p> <p>You need not demand the second square to be commutative, it automatically commutes by uniqueness of adjoints. The wise choice is:</p> <p>$$f_\ast k_\ast \stackrel{f_\ast k_\ast\eta_h}\longrightarrow f_\ast k_\ast h^\ast h_\ast \stackrel{f_\ast k_\ast h^\ast \eta_g h_\ast}\longrightarrow f_\ast k_\ast h^\ast g^\ast g_\ast h_\ast \stackrel{f_\ast k_\ast \zeta^{-1} g_\ast h_\ast}\longrightarrow f_\ast k_\ast k^\ast f^\ast g_\ast h_\ast \stackrel{f_\ast \epsilon_k f^\ast g_\ast h_\ast}\longrightarrow f_\ast f^\ast g_\ast h_\ast \stackrel{\epsilon_f g_\ast h_\ast}\longrightarrow g_\ast h_\ast$$</p> <p>This choice is precisely taken from the usual proof of uniqueness of adjoints.</p> <p>With this choice, it's a straightforward exercise on 2-categorical composition to check that your third diagram commutes. You must also use the equations satisfied by (co)units in an adjunction.</p> http://mathoverflow.net/questions/23269/non-examples-of-model-structures-that-fail-for-subtle-surprising-reasons/106949#106949 Answer by Fernando Muro for Non-examples of model structures, that fail for subtle/surprising reasons? Fernando Muro 2012-09-11T19:53:53Z 2013-06-08T22:20:29Z <p>I like the following example because it is very close to the origins of homotopy theory (and also because I worked on it at the beginning of my career): <em>proper homotopy theory</em>. Objects are topological spaces, maps are proper maps, one can define proper homotopies via cylinders in the usual way, weak equivalences are proper homotopy equivalences, and cofibrations are proper maps satisfying the homotopy extension property. The 'cofibrant' part works as in a model category, this is a cofibration category, but there are very few fibrations. Moreover, the category is not complete, it doesn't even have a final object. This is because the map to a point $X\rightarrow *$ is not proper unless $X$ is compact. Proper homotopy theory is however very much developed on its own, and has been applied in many contexts.</p> http://mathoverflow.net/questions/131345/why-every-complex-of-injectives-is-homotopically-injective-provided-that-the-in/131360#131360 Answer by Fernando Muro for Why every complex of injectives is homotopically injective (provided that, the injective dimension is finite)? Fernando Muro 2013-05-21T16:30:46Z 2013-05-21T16:30:46Z <p>Let $N$ be an acyclic complex of injectives. I will denote complexes with cohomological degre, i.e. degree $+1$ differentials. Notice that the complex $N$ is contractible iff it the kernel of each differential $d^k\colon N^k \rightarrow N^{k+1}$ is injective. Suppose that $\mathscr A$ has finite injective dimension $n$. For any <code>$l&lt;k$</code> we have an exact sequence</p> <p>$$\ker d^l\hookrightarrow N^l\rightarrow N^{l+1}\rightarrow\cdots\rightarrow N^{k-1}\twoheadrightarrow \ker d^k$$</p> <p>This is in particular true for $k-l>n$. By hypothesis, $\ker d^{l+n}$ is injective, hence the exact sequence $$\ker d^{l+n}\hookrightarrow N^{l+n}\rightarrow N^{l+n+1}\rightarrow\cdots\rightarrow N^{k-1}\twoheadrightarrow \ker d^k$$</p> <p>shows that $\ker d^k$ must also be injective.</p> http://mathoverflow.net/questions/129494/is-this-square-a-push-out-square/129511#129511 Answer by Fernando Muro for Is this square a push-out square? Fernando Muro 2013-05-03T10:14:17Z 2013-05-03T14:27:12Z <p>I guess you mean 'isomorphism' where you say 'equivalence' of $R$-modules. With your assumptions it is very easy to check (diagram chasing) that the folded sequence $$0\rightarrow B\stackrel{\binom{q}{?}}\longrightarrow C\oplus E\stackrel{(?,-r)}\longrightarrow F\rightarrow 0$$ is short exact. Hence the square on the right is both a pull-back and a push-out, essentially by definition. In particular, since it is a pull-back, the kernels of the two vertical maps on the right are the same.</p> http://mathoverflow.net/questions/129353/example-re-torsionless-quotients-of-abelian-groups/129356#129356 Answer by Fernando Muro for example re torsionless quotients of abelian groups Fernando Muro 2013-05-01T22:26:57Z 2013-05-01T22:33:11Z <p>Take $B$ any torsionless abelian group, $T$ any abelian group, $S=T\oplus \mathbb Z/2$ and $A=B\oplus S$, then $A/T=B\oplus\mathbb Z/2$ is not torsionless.</p> http://mathoverflow.net/questions/127139/homotopy-excision-and-homotopy-pushout/127147#127147 Answer by Fernando Muro for Homotopy excision and homotopy pushout Fernando Muro 2013-04-10T20:55:46Z 2013-04-10T20:55:46Z <p>1) I assume your three model structures have the same weak equivalences, correct me if I'm wrong. Let $\mathcal C$ be a model category and $I$ a small category, e.g. $I=\bullet\leftarrow \bullet\rightarrow\bullet$ if you're interested in push-outs. The homotopy colimit functor $\operatorname{hocolim}_I\colon\operatorname{Ho}(\mathcal C^I)\rightarrow \operatorname{Ho}(\mathcal C)$ is simply the left adjoint to the constant functor $\operatorname{Ho}(\mathcal C)\rightarrow\operatorname{Ho}(\mathcal C^I)$. Homotopy categories only depend on weak equivalences, hence homotopy colimits too.</p> <p>2 &amp; 3) Yes, it is a beautiful result in:</p> <p>MR1452856 Chachólski, Wojciech A generalization of the triad theorem of Blakers-Massey. Topology 36 (1997), no. 6, 1381–1400.</p> http://mathoverflow.net/questions/126938/commutativity-of-tor/126939#126939 Answer by Fernando Muro for Commutativity of Tor Fernando Muro 2013-04-09T06:23:39Z 2013-04-09T10:25:06Z <p>Yes, the derived category of a commutative ring is symmetric monoidal $M\otimes_A^{\mathbb L}N=N\otimes_A^{\mathbb L}M$ and Tor is the homology of this tensor product.</p> http://mathoverflow.net/questions/125773/a-statement-for-a-subset-generated-a-triangulated-category/125777#125777 Answer by Fernando Muro for A statement for a subset generated a triangulated category Fernando Muro 2013-03-28T01:27:08Z 2013-03-28T01:34:26Z <p>$\Leftarrow$ is false: take $D=D^b(\mathbb Z)$ the derived category of bounded complexes of abelian groups and $A=\{\mathbb Z^2\}$. Then $\mathbb Z\notin \langle A\rangle$ but $A^{\perp}=0$. Indeed, $\langle A\rangle$ is formed by bounded complexes of free abelian groups of finite even rank, hence by Euler characteristic arguments $\mathbb Z\notin \langle A\rangle$.</p> http://mathoverflow.net/questions/125125/categorical-description-of-the-second-k-group/125137#125137 Answer by Fernando Muro for Categorical description of the second K-group Fernando Muro 2013-03-21T08:21:24Z 2013-03-21T08:21:24Z <p>The $K_1$ group you describe is the automorphism $K_1$, which is in general <em>not</em> isomorphic to Quillen's $K_1$ of an exact category. It coincides with Quillen's when exact sequences in $\mathcal P$ split. For a description by generators and relations of $K_1$ of any exact category see:</p> <p>Nenashev, A. K1 by generators and relations. (English summary) J. Pure Appl. Algebra 131 (1998), no. 2, 195–212. </p> <p>generators are pairs of short exact sequences on the same objects, and relations are given by $3\times 3$ diagrams. A generalization of this result to all $K_n$ is given in</p> <p>Grayson, Daniel R. Algebraic K-theory via binary complexes. J. Amer. Math. Soc. 25 (2012), no. 4, 1149–1167. </p> http://mathoverflow.net/questions/123967/bases-of-open-sets-with-connected-intersections Bases of open sets with connected intersections Fernando Muro 2013-03-08T11:56:40Z 2013-03-08T13:44:50Z <p>I'm interested in knowing classes of topological spaces $X$ which admit a basis of open sets $\{U_i\}_{i\in I}$ such that $U_i\cap U_j$ is connected for all $i,j\in I$. Do manifolds have this property? Riemannian ones maybe? If not, what if we relax the condition by saying that $U_i\cap U_j$ has a finite number of connected components for $i,j\in I$? </p> http://mathoverflow.net/questions/123966/constants-sheaves-on-an-open-subset Constants sheaves on an open subset Fernando Muro 2013-03-08T11:53:15Z 2013-03-08T13:29:23Z <p>Let $X$ be a topological space and $U\subset X$ an open subset. Let's work in the category of sheaves of abelian groups on $X$. Consider the constant sheaf on $U$, $\mathbb{Z}_U$, given by $\mathbb{Z}_U(V)=\{\text{contonuous maps }U\cap V\rightarrow \mathbb Z\}$, where $\mathbb Z$ is given the discrete topology. I've been struggling to derive from Yoneda's lemma the formula $\hom(\mathbb{Z}_U,F)=F(U)$. Is this a consequence of Yoneda? If so, how? If it doesn't follow from Yoneda, is it true at all? If not, how can one compute $\hom(\mathbb{Z}_U,\mathbb{Z}_V)?$</p> http://mathoverflow.net/questions/123283/morphisms-between-k-0/123304#123304 Answer by Fernando Muro for Morphisms between $K_0$ Fernando Muro 2013-03-01T07:26:56Z 2013-03-02T00:08:17Z <p>I'd say no to both questions. </p> <p>1) If $k$ is a field then $K_0(k)= \mathbb Z$ generated by $[k]$ and the class of any $k$-module is positive, so $-n\colon K_0(k)\rightarrow K_0(k)$ cannot be induced by a bimodule, $n>0$. </p> <p>2) If $k$ is a field of positive characteristic and $k'$ is a field of characteristic $0$ then the only map $K_0(k)\rightarrow K_0(k')$ induced by a bimodule is the trivial map, since the only $k$-$k'$-bimodule is the trivial one. The same the other way round.</p> <p>EDIT: Answering Sasha's comment below. If $A=\mathbb{C}[\epsilon]/(\epsilon^2)$, any left and right projective $k$-$A$-bimodule is even-dimensional over $\mathbb C$, hence all induced homomorphisms $K_0(A)=\mathbb Z\rightarrow K_0(\mathbb C) =\mathbb Z$ are multiples of $2$, in particular the identity fails to be induced.</p> http://mathoverflow.net/questions/123358/when-is-this-braiding-not-a-symmetry/123380#123380 Answer by Fernando Muro for When is this braiding not a symmetry? Fernando Muro 2013-03-01T22:58:31Z 2013-03-01T22:58:31Z <p>If $X=\Omega Y$, that braided monoidal category (indeed groupoid) classifies the homotopy type of $P_3Y$, the $3$-type of $Y$. Such $3$-type is completely determined by the map $\eta^*\colon \pi_2(Y)\rightarrow \pi_3(Y)$ defined by precomposition with the Hopf map $\eta\colon S^3\rightarrow S^2$. This map is quadratic, i.e. $\eta^* (x)=\eta^* (-x)$ and the map defined by $\eta^* (x|y)=\eta^* (x+y)-\eta^* (x)-\eta^* (y)$ is bilinear. One can recover this map from the monoidal category, essentially $\eta^*(x)$ corresponds to the braiding $\gamma_{x,x}\colon x\otimes x\cong x\otimes x$. The category is symmetric if and only if $\eta^*$ is a homomorphism. This does not always happen, since any quadratic map between abelian groups $A\rightarrow B$ can be realized by some appropriate $Y$. For instance, if you take $Y=S^2$ the quadratic map is $\eta^* \colon \mathbb{Z}\rightarrow \mathbb{Z}$ is $\eta^* (n)=n^2$, hence you get an example.</p> http://mathoverflow.net/questions/121140/universality-of-ext-functor-using-yoneda-extensions/121957#121957 Answer by Fernando Muro for Universality of Ext functor using Yoneda extensions Fernando Muro 2013-02-15T23:26:25Z 2013-02-15T23:26:25Z <p>In their paper entitled "Extension categories and their homotopy", Neeman and Retakh define a spectrum of extensions $\operatorname{Ext}(A,B)$ for any two objects in an exact category $\mathcal E$ such that $\pi_{-n}\operatorname{Ext}(A,B)=\operatorname{Ext}_{\mathcal E}^n(A,B)$, in the sense of Yoneda, for any $n\geq 0$. Positive-dimensional homotopy groups vanish. The spectrum $\operatorname{Ext}(A,B)$ is an $\Omega$-spectrum defined by the classifying spaces of the categories $\operatorname{Ext}^n(A,B)$ of $n$-fold Yoneda extensions. </p> <p>Given a short exact sequence $B\hookrightarrow C\twoheadrightarrow D$, Quillen's Theorem B shows that the homotopy fiber of $\operatorname{Ext}^n(A,C)\rightarrow \operatorname{Ext}^n(A,D)$ is $\operatorname{Ext}^n(A,B)$, $n\geq 0$. Hence, for spectra, the homotopy fiber of $\operatorname{Ext}(A,C)\rightarrow \operatorname{Ext}(A,D)$ is $\operatorname{Ext}(A,B)$. The long exact sequence on homotopy groups defines now a $\delta$-functor $\operatorname{Ext}^\bullet(A,-)$.</p> <p>Universality follows from Yoneda's lemma. If $T$ is another $\delta$-functor, a natural transformation $\operatorname{Hom}(A,-)=\operatorname{Ext}^0(A,-)\rightarrow T^0$ extends uniquely to a morphism of $\delta$-functors $\operatorname{Ext}^n(A,-)\rightarrow T^n$, $n\geq 0$, as follows. An $n$-fold extension $B\hookrightarrow X_1\rightarrow\cdots\rightarrow X_n\twoheadrightarrow A$ factors as the 'composition' of short exact sequences $$Y_{n-1}\hookrightarrow X_n\twoheadrightarrow Y_n$$ with $Y_0=B$ and $Y_n=A$. In particular we obtain morphisms $$T^0(A)\rightarrow T^1(Y_{n-1})\rightarrow T^2(Y_{n-2})\rightarrow\cdots\rightarrow T^{n-1}(Y_1)\rightarrow T^n(B).$$ The image of the previous extension by extension by $\operatorname{Ext}^n(A,B)\rightarrow T^n(B)$ is the image by this composite of the element in $T^0(A)$ classifying the natural transformation we started with (via Yoneda's lemma). Everything is well defined by the properties defining a $\delta$-functor.</p> <p>This is only a sketch of proof. If you intend to use it in a paper you should probably provide some details at some points, e.g. carefully check the hypotheses of Quillen's Theorem B.</p> http://mathoverflow.net/questions/120875/ring-with-three-binary-operations/120900#120900 Answer by Fernando Muro for Ring with three binary operations Fernando Muro 2013-02-05T19:08:57Z 2013-02-05T21:50:55Z <p>The real numbers $\mathbb{R}$ with the following three binary operations:</p> <ul> <li><p>The maximum: $(x,y)\mapsto\max\{x,y\}$.</p></li> <li><p>The sum: $(x,y)\mapsto x+y$.</p></li> <li><p>The product: $(x,y)\mapsto x\cdot y$.</p></li> </ul> <p>The maximum is to the sum what the sum is to the product, except for the fact that the maximum does not have inverses, nor a unit, i.e. $(\mathbb{R},\max,+)$ is a semiring, while $(\mathbb{R},+,\cdot)$ is a ring.</p> http://mathoverflow.net/questions/120433/groups-lying-horizontally-in-2-groups/120462#120462 Answer by Fernando Muro for Groups lying horizontally in 2-groups Fernando Muro 2013-01-31T21:03:11Z 2013-01-31T21:03:11Z <p>Let $\partial\colon C_1\rightarrow C_0$ be a crossed module, i.e. a group homomorphism together with a right action of $C_0$ on $C_1$, denoted exponentially $c_1^{c_0}$, satisfying the following laws: $$\partial(c_1^{c_0})=-c_0+c_1+c_0$$ $$c_1^{\partial(c_1')}=-c_1'+c_1+c_1'.$$ Here I denote the group laws additively, despite they are non abelian in general.</p> <p>Crossed modules are equivalent to strict $2$-groups. The strict $2$-group $C_*$ associated to $\partial\colon C_1\rightarrow C_0$ has set of objects in $C_0$ and set of morphisms $C_0\ltimes C_1$, the semidirect product. I'll show that the group you define for $C_*$ is this semidirect product $C_0\ltimes C_1$. Hence it is <strong>not invariant under tensor equivalences</strong>.</p> <p>The source and target of $(c_0,c_1)\in C_0\ltimes C_1$ are $$(c_0,c_1)\colon c_0+\partial(c_1)\longrightarrow c_0.$$ Composition is defined by $$(c_0,c_1)\circ(c_0+\partial(c_1),c_1')=(c_0,c_1+c_1').$$ The tensor product is the sum $+$ and the tensor unit is $0\in C_0$. In this case $\epsilon$ and $\eta$ are the identity map on the tensor unit, i.e. $(0,0)\in C_0\ltimes C_1$. </p> <p>The mate of a morphism $f=(c_0,c_1)\in C_0\ltimes C_1$ is $\hat{f}=(-c_0,-c_1^{-c_0})\in C_0\ltimes C_1$. The only simple object is the tensor unit $0\in C_0$ and two morphisms $f=(c_0,c_1),g=(c_0',c_1')\in C_0\ltimes C_1$ are quivalent iff $$(0,0)=f+\hat{g}=(c_0,c_1)+(-c_0',-(c_1')^{-c_0'})=(c_0-c_0',c_1^{-c_0'}-(c_1')^{-c_0'}) =(c_0-c_0',(c_1-c_1')^{-c_0'})$$ i.e. iff $c_0=c_0'$ and $c_1=c_1'$, that is $f=g$. Therefore your group is simply the semidirect product $C_0\ltimes C_1$.</p> http://mathoverflow.net/questions/120235/is-the-derived-category-of-abelian-groups-a-subcategory-of-the-stable-homotopy-ca/120268#120268 Answer by Fernando Muro for Is the derived category of abelian groups a subcategory of the stable homotopy category? Fernando Muro 2013-01-29T22:42:50Z 2013-01-29T22:42:50Z <p>I've found the following somewhat intricate way of answering Q1 in the affirmative. Any complex in $D(Ab)$ quasi-isomorphic to a graded abelian group. Hence, it is enough to consider complexes concentrated in a single degree. Given an abelian group $A$ and $n\in\mathbb Z$, let $(A,n)$ be the abelian group $A$ concentrated in degree $n$. For simplicity, I will use the same notation for the Eilenberg-MacLane spectrum $\Sigma^nHA$. In the derived category we have, $$D(Ab)((A,n),(B,n))=\operatorname{Hom}(A,B),$$ $$D(Ab)((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$ $$D(Ab)((A,n),(B,m))=0\text{ otherwise}.$$ In the stable homotopy category we have the stable Eilenberg-MacLane groups $$SH((A,n),(B,m))=H^{m+k}(A,n+k;B),\quad k>>0.$$ It is well known, since E-ML's "On the groups..." (Annals) that $$SH((A,n),(B,n))=\operatorname{Hom}(A,B),$$ $$SH((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$ and that the functor $D(Ab)\rightarrow SH$ is the identity on the previous morphism sets. Hence we are done. The groups $SH((A,n),(B,m))$ are however non-trivial for $m>n+1$, in general.</p> http://mathoverflow.net/questions/119475/example-a-pair-of-nonisomorphic-parallel-morphisms-with-isomorphic-cones/119546#119546 Answer by Fernando Muro for Example: a pair of nonisomorphic parallel morphisms with isomorphic cones Fernando Muro 2013-01-22T07:54:07Z 2013-01-22T07:54:07Z <p>Yet another example. Take $R$ any ring such that $R\cong R\oplus R$. Consider the following parallel morphisms $f,g\colon R\rightarrow R$: $f=0$ the trivial morphism, and</p> <p>$$g=\left(\begin{smallmatrix} 1&amp;0\\0&amp;0 \end{smallmatrix}\right)\colon R\cong R\oplus R\longrightarrow R\oplus R\cong R.$$ Both have isomorphic mapping cone</p> <p>$$C\colon \cdots\rightarrow0\rightarrow R\stackrel{0}\rightarrow R\rightarrow0\rightarrow\cdots$$</p> <p>but $f\ncong g$ since $f=0\neq g$.</p> http://mathoverflow.net/questions/118058/irreducible-cohomology-theories/118072#118072 Answer by Fernando Muro for Irreducible cohomology theories Fernando Muro 2013-01-04T17:34:34Z 2013-01-04T17:34:34Z <p>A generalized cohomology theory is indecomposable iff its classifying spectrum is. This follows from Brown representability. Baues and Drozd have a paper in Topology about finite-dimensional indecomposable stable homotopy types withney fg free homology. In diension 4, you have 3 indecomposable spectra with rank 1 homology in dimensions 0 and 4, rank 2 homology in dimension 2, and 0 elsewhere, see definition 1.7. Whitehead's long exact sequence for the Hurewicz homomorphism looks as follows towards the end:</p> <p>$$\pi_2\rightarrow H_2=\mathbb Z^2 \rightarrow \Gamma_1=H_0\otimes \mathbb Z/2=\mathbb Z/2$$</p> <p>This implies that the second generalized homology of the point wrt this spectrum, which is $\pi_2$, contains a subgroup isomorphic to $\mathbb Z^2$.</p> http://mathoverflow.net/questions/118044/an-isomorphism-between-different-exts-coming-from-group-cohomology/118049#118049 Answer by Fernando Muro for An isomorphism between different Ext's coming from group cohomology Fernando Muro 2013-01-04T12:25:49Z 2013-01-04T12:25:49Z <p>The group $H^2(G,M)$ classifies different types of extensions than $\operatorname{Ext}^1(G,M)$. On the one hand, $H^2(G,M)$ classifies extensions</p> <p>$$M\hookrightarrow H\twoheadrightarrow G$$</p> <p>where $H$ may be non-abelian, and the action of $H$ on $M$ by conjugation is encoded in the $G$-module structure of $M$. On the other hand, $\operatorname{Ext}^1(G,M)$ classifies extensions</p> <p>$$M\hookrightarrow A\twoheadrightarrow G$$</p> <p>where $A$ is an abelian group, in particular $A$ acts trivially on $M$ by conjugation, i.e. $M$ can only be regarded as a trivial $G$-module here.</p> <p>Even if you regard $M$ as a trivial $G$-module in both cases, $H^2(G,M)$ and $\operatorname{Ext}^1(G,M)$ may be different due to the existence of non-abelian but central extensions. In general, there is a universal coefficient split short exact sequence</p> <p>$$\operatorname{Ext}^1(G,M)\hookrightarrow H^2(G,M)\twoheadrightarrow \operatorname{Hom}(H_2G,M).$$</p> <p>You can find this in most books on group cohomology. The first morphism represents the inclusion of abelian extensions into central (but possibly non-abelian) extensions (recall that $M$ carries here the trivial $G$-module structure). The group $\operatorname{Hom}(H_2G,M)$ measures the amount of really non-abelian central extensions of $G$ by $M$.</p> <p>Fortunately, $H_2G$ is very easy to compute, it is the exterior square $H_2G=\wedge^2G$, i.e. the quotient of $G\otimes G$ by the relations $g\otimes g=0$, $g\in G$. This functor is quadratic, $$\wedge^2(G_1\oplus G_2)=\wedge^2(G_1)\oplus (G_1\otimes G_2) \oplus \wedge^2(G_2)$$ and vanishes on (finite or infinite) ciclyc groups $\wedge^2(\mathbb{Z}/n)=0$, $n\in\mathbb Z$. This gives a recipe to compute $H_2G$ for any finitely generated abelian group. In particular, if you take $G=(\mathbb{Z}/2)^2$ and $M=\mathbb{Z}/2$ you get</p> <p>$$\operatorname{Ext}^1(G,M)=(\mathbb{Z}/2)^2,\qquad \operatorname{Hom}(H_2G,M)=\mathbb Z/2.$$ Hence $$H^2(G,M)=(\mathbb{Z}/2)^3.$$</p> http://mathoverflow.net/questions/117986/a-lost-lemma-about-periodicity-in-a-grid-of-long-exact-sequences/117991#117991 Answer by Fernando Muro for A lost lemma about periodicity in a grid of long exact sequences? Fernando Muro 2013-01-03T20:32:15Z 2013-01-03T20:32:15Z <p>Everything can be reduced to long exact sequences induced by short exact sequences of complexes.</p> <p>In your setting, there are short exact sequences of complexes as follows</p> <p>$$0\rightarrow C_{11}\stackrel{(u,g)}\longrightarrow C_{12}\oplus C_{21}\longrightarrow C_{12}\cup_{C_{11}}C_{21}\rightarrow 0$$</p> <p>$$0\rightarrow C_{12}\cup_{C_{11}}C_{21}\stackrel{(g,-u)}\longrightarrow C_{22}\stackrel{h\nu}\longrightarrow C_{33}\rightarrow 0$$</p> <p>This produces long exact sequences</p> <p>$$\cdots\rightarrow H^{k}C_{11}\longrightarrow H^{k}C_{12}\oplus H^{k}C_{21}\longrightarrow H^{k}(C_{12}\cup_{C_{11}}C_{21})\longrightarrow H^{k+1}C_{11}\rightarrow \cdots$$</p> <p>$$\cdots\rightarrow H^{k}(C_{12}\cup_{C_{11}}C_{21})\longrightarrow H^{k}C_{22}\longrightarrow H^{k}C_{33}\longrightarrow H^{k+1}(C_{12}\cup_{C_{11}}C_{21})\rightarrow \cdots$$</p> <p>Your hypotheses say that </p> <p>$$H^{k}C_{12}\oplus H^{k}C_{21}\longrightarrow H^{k}(C_{12}\cup_{C_{11}}C_{21})\longrightarrow H^{k}C_{22}$$</p> <p>$$([\alpha],[\beta])\mapsto [\alpha-\beta]\mapsto 0$$</p> <p>therefore there exists $[\gamma]\in H^{k-1}(C_{33})$ such that</p> <p>$$H^{k-1}C_{33}\longrightarrow H^{k}(C_{12}\cup_{C_{11}}C_{21})$$</p> <p>$$[\gamma]\mapsto [\alpha-\beta]$$</p> <p>Now it is enough to compose with the morphism induced in cohomology by</p> <p>$$\left(\begin{smallmatrix}\nu&amp;0\\0&amp;h\end{smallmatrix}\right)\colon C_{12}\cup_{C_{11}}C_{21}\longrightarrow C_{13}\oplus C_{31}$$</p> <p>in order to obtain the thesis of your lemma. (BTW, notice that there is a misprint in your subscripts, you must replace two 2s by 1s)</p> http://mathoverflow.net/questions/117886/recovering-torsion-in-singular-homology-from-cplx-of-singular-chains/117887#117887 Answer by Fernando Muro for Recovering torsion in singular homology from cplx of singular chains Fernando Muro 2013-01-02T18:51:51Z 2013-01-02T19:10:42Z <p>1) is false, Whitehead's theorem doesn't say that. Actually, any complex over a hereditary ring, eg the integers, is quasi-isomorphic to its cohomology. </p> <p>2) No, lens spaces have quasi-isomorphic singular (co)chains but different integral cohomology. </p> <p>3) Yes, by the answer to 1)</p> <p>Maybe you're interested in doing all this <em>functorially</em>. Since this is a very important point, if this is what you want you should specify all this explicitly, eg what would be the source category, the target, etc.</p> http://mathoverflow.net/questions/117857/exact-sequences/117858#117858 Answer by Fernando Muro for Exact sequences Fernando Muro 2013-01-02T11:13:52Z 2013-01-02T11:13:52Z <p>Not in general. The keyword is stable module category, the quotient of the module category by the ideal of morphisms which factor through a projective. The leftmost term is functorial on the rightmost term in this category if all intermediate modules are projective. This imposes some restrictions. If you take a hereditary ring, eg the integers, you get easy counterexamples as any submodule of a projective module is projective.</p> http://mathoverflow.net/questions/116975/are-filtered-colimits-of-weak-equivalences-of-spectra-again-weak-equivalences/116987#116987 Answer by Fernando Muro for Are filtered colimits of weak-equivalences of spectra again weak-equivalences? Fernando Muro 2012-12-21T19:15:12Z 2012-12-21T19:15:12Z <p>The answer is <strong>yes</strong>, but the reason is technical. </p> <p>The reason is that, if I understand well, you're asking whether weak equivalences are closed under <em>arbitrary</em> filtered colimits. These are $\aleph_0$-filtered colimits, and there is a hierarchy of degrees of filtration parametrized by all infinite regular cardinals $\alpha$, being $\alpha=\aleph_0$ the first one. </p> <p>In general combinatorial model categories, you know that there exists a big enough regular cardinal $\lambda$ such weak equivalences are closed under $\alpha$-filtered colimits for all $\alpha\geq \lambda$, but not necessarily for $\alpha&lt;\lambda$. </p> <p>The category of spectra of simplicial sets is locally $\aleph_0$-presentable and has sets of generating (trivial) cofibrations with $\aleph_0$-presentable sources. Hence you can take $\lambda=\aleph_0$.</p> <p>See section 7 of</p> <p>D. Dugger, Combinatorial model categories have presentations, Adv. Math. 164 (2001), 177–201,</p> <p>and the appendix of</p> <p>S. Schwede, Stable homotopy of algebraic theories, Topology 40 (2001), 1–41.</p> http://mathoverflow.net/questions/114272/chain-condition-on-rings/114276#114276 Answer by Fernando Muro for Chain Condition on Rings Fernando Muro 2012-11-23T18:35:48Z 2012-11-23T18:35:48Z <p>Take the Weil algebra over a field $k$, i.e. $R=k\langle x,\frac{d}{dx}\rangle$ is the algebra with two generators, $x$ and $\frac{d}{dx}$, and one relation $\frac{d}{dx} x=1$. This algebra is well known to be simple, so it satisfies the DCC on two-sided ideals. For right ideals you have $xR\supsetneq x^2R\supsetneq x^3R\supsetneq\cdots$, and for left ideals $R\frac{d}{dx}\supsetneq R(\frac{d}{dx})^2\supsetneq R(\frac{d}{dx})^3\supsetneq\cdots$.</p> http://mathoverflow.net/questions/114241/projective-arrows/114247#114247 Answer by Fernando Muro for Projective arrows Fernando Muro 2012-11-23T15:05:07Z 2012-11-23T18:11:30Z <p>If you assume that your category $\mathcal C$ has enough projectives, e.g. the category of groups, then your projective arrows are the maps which factor through a projective. Let us check this. </p> <p>Let $f\colon X\rightarrow Y$ be a projective arrow. Since $\mathcal C$ has enough projectives, we can take an epimorphism $g\colon P\twoheadrightarrow Y$ with projective source, hence $f$ factors through $P$. Conversely, if $f\colon X\rightarrow Y$ factors as</p> <p>$$f\colon X\stackrel{f''}\rightarrow Q\stackrel{f'}\rightarrow Y$$</p> <p>with $Q$ projective, then $f'$ can be lifted along any epimorphism $g\colon P\twoheadrightarrow Y$, hence $f$ too.</p> <p>Maps which factor through a projective are widely studied in homological algebra. The stable category $\underline{\mathcal A}$ of an abelian category $\mathcal A$ is the quotient of $\mathcal A$ by the ideal of morphisms factoring through a projective, i.e. by the ideal of projective maps in your terminology. You can read in wikipedia about the stable category of modules over a ring:</p> <p><a href="http://en.wikipedia.org/wiki/Stable_module_category" rel="nofollow">http://en.wikipedia.org/wiki/Stable_module_category</a></p> <p>if ${\mathcal A}$ is a Frobenius abelian category, i.e. enough projectives and injectives and both classes of objects coincide, then $\underline{\mathcal A}$ is a triangulated category. All algebraic triangulated categories arise in this way, but allowing ${\mathcal A}$ to be an exact category, not just abelian.</p> http://mathoverflow.net/questions/110658/fubini-theorem-for-hocolim/110683#110683 Answer by Fernando Muro for Fubini theorem for hocolim. Fernando Muro 2012-10-25T15:41:10Z 2012-10-25T15:41:10Z <p>This property holds actually for right derivable categories in the sense of: </p> <p>MR2729017 Reviewed Cisinski, Denis-Charles Catégories dérivables. (French) [Derivable categories] Bull. Soc. Math. France 138 (2010), no. 3, 317–393.</p> <p>At least under suitable finiteness assumptions on $I$ and $J$. It also holds for arbitrary small categories $I$ and $J$ when working on a homotopically complete right derivable category. Model categories are examples of these (cofibrant generation is not needed).</p> <p>Cisinski shows that the homotopy categories of diagrams on a (homotopically complete) right derivable category $\mathcal{C}$ form a right derivator whose domain is the category of directed finite categories (or all small categories in the homotopically complete case). In particular, one of the axiom says that if $u\colon A\rightarrow B$ is any functor in the domain and $u_!\colon\operatorname{Ho}(\mathcal{C}^A)\rightarrow \operatorname{Ho}(\mathcal{C}^B)$ is the right adjoint of the restriction along $u$ functor $u^{*}\colon\operatorname{Ho}(\mathcal{C}^B)\rightarrow \operatorname{Ho}(\mathcal{C}^A)$ then for any $F\colon A^{\operatorname{op}}\rightarrow \mathcal{C}$ and any object $b\in B$ the formula $u_!(F)(b)=p_!(A\downarrow b\rightarrow A\stackrel{F}\rightarrow \mathcal{C})$ holds. Here $p_!\colon\operatorname{Ho}(\mathcal{C}^{A\downarrow b})\rightarrow \operatorname{Ho}(\mathcal{C})$ is the usual homotopy colimit.</p> <p>In your case, take $u\colon I\times J\rightarrow I$ to be the projection onto the first factor. Then <code>$(\operatorname{hocolim}_JF)(i)=u_{!}(F)(i)$</code> by definition, the functor $J\rightarrow(I\times J)\downarrow i\colon j\mapsto (i,j)$ is cofinal, and hence <code>$(\operatorname{hocolim}_JF(i))=p_!((I\times J)\downarrow i\rightarrow I\times J\stackrel{F}\rightarrow \mathcal{C})$</code>.</p> http://mathoverflow.net/questions/110444/determining-homotopy-classes-t2-rp2/110541#110541 Answer by Fernando Muro for Determining homotopy classes [T^2, RP^2] Fernando Muro 2012-10-24T13:57:50Z 2012-10-24T21:11:22Z <p>For dimension reasons, the set of homotopy classes of maps between these spaces can be computed by using their <em>fundamental crossed module</em>. Despite the question has already been answered twice, let me explain this approach, just for those who may like it.</p> <p>A <em>crossed module</em> $C_*$ is a group homomorphism $\partial\colon C_2\rightarrow C_1$ together with an action of $C_1$ on $C_2$, that we write exponentially $c_2^{c_1}$, such that </p> <p>$$\begin{array}{rl} \partial(c_2^{c_1})&amp;=-c_1+\partial(c_2)+c_1,\\ c_2^{\partial(c_2')}&amp;=-c_2'+c_2+c_2'. \end{array}$$ Here we use additive notation despite the groups may be nonabelian.</p> <p>The canonical (topological) example of a crossed module is the boundary map in the long exact sequence of a pair of spaces:</p> <p>$$\partial\colon\pi_2(X,Y)\longrightarrow\pi_1(X).$$</p> <p>The <em>fundamental crossed module</em> of CW-complex $X$ is obtained in this way for the pair formed by $X$ and its $1$-skeleton.</p> <p>$$\partial\colon\pi_2(X,X^1)\longrightarrow\pi_1(X^1).$$</p> <p>Notice that $\ker\partial\cong\pi_2(X)$. The image of $\partial$ is normal in any crossed module, and in this case $\operatorname{coker} \partial\cong\pi_1(X)$.</p> <p>A <em>morphism</em> of crossed modules $f_{*}\colon C_{*}\rightarrow D_{*}$ is a commutative square</p> <p>$$\begin{array}{rcccl} &amp;C_1&amp;\stackrel{\partial}\longrightarrow&amp;C_2\\ {\scriptstyle f_2}&amp;\downarrow&amp;&amp;\downarrow&amp;{\scriptstyle f_1}\\ &amp;D_1&amp;\stackrel{\partial}\longrightarrow&amp;D_2 \end{array}$$</p> <p>such that $f_2(c_2^{c_1})=f_2(c_2)^{f_1(c_1)}$. Notice that any cellular map between CW-complexes $X\rightarrow Y$ induces a morphism between their fundamental crossed modules.</p> <p>Two such morphisms $f_{*},g_{*}\colon C_{*}\rightarrow D_{*}$ are <em>homotopic</em> if there exists a map $H\colon C_1\rightarrow D_2$ such that $$\begin{array}{rl} H(c_1+c_1')&amp;=H(c_1)^{f(c_1')}+H(c_1'),\\ \partial H(c_1)&amp;=-f_1(c_1)+g_1(c_1),\\ H\partial(c_2)&amp;=-f_2(c_2)+g_2(c_2). \end{array}$$</p> <p>If $X$ and $Y$ are CW-complexes and $X$ is $2$-dimensional the set of homotopy classes $[X,Y]$ can be computed as the set of algebraic homotopy classes between their fundamental crossed modules.</p> <p>The fundamental crossed module of $T^2$ is isomorphic to $\partial\colon G'\hookrightarrow G=\langle a,b\rangle$. Here $G$ os a free group on two generators, $G'$ is its commutator subgroup, and $\partial$ is the inclusion. The action of $G$ on $G'$ is by conjugation.</p> <p>The fundamental crossed module of $\mathbb{R}P^2$ is even easier: $\partial=2\cdot\varepsilon\colon \mathbb{Z}[\mathbb{Z}/2]\rightarrow \mathbb{Z}$. Here $\varepsilon\colon \mathbb{Z}[\mathbb{Z}/2]\rightarrow \mathbb{Z}$ is the augmentation map of the group ring and $\mathbb{Z}$ acts on $\mathbb{Z}[\mathbb{Z}/2]$ via the natural projection $\mathbb{Z}\twoheadrightarrow \mathbb{Z}/2$.</p> <p>It is a beautiful exercise to compute the set of homotopy classes of maps between these two crossed modules. It's easy since the second one is made of abelian groups! If you're given a map $T^2\rightarrow\mathbb{R}P^2$ and you manage to find a homotopic cellular map (e.g. using the proof of the cellular approximation theorem), then you can say what homotopy class you started with by looking at the induced morphism on the level of crossed modules.</p> http://mathoverflow.net/questions/108737/small-categories-and-completeness/108740#108740 Answer by Fernando Muro for Small categories and completeness Fernando Muro 2012-10-03T19:32:36Z 2012-10-03T19:32:36Z <p>(1) Yes, the trivial (final) category, with only one object and one morphism (the identity in the unique object).</p> <p>(2) If you only asked for colimits of cardinality $&lt;\kappa$ then you would find many examples, e.g. finite abelian groups ($\kappa = \aleph_0$ here). If you insists in $\leq\kappa$ I think you face the same problem as above. Just think of vector spaces of dimension $&lt;\kappa$. This category, up to isomorphism, has $\leq \kappa$ objects but it doesn't have colimits of sice $\leq \kappa$ since the coproduct of $\kappa$ copies of the ground field has dimension exactly $\kappa$.</p> <p>I've been speaking about colimits in (2) instead of limits, which is what you ask for, so take opposite categories.</p> http://mathoverflow.net/questions/91510/whitehead-product-with-identity-on-homotopy-groups-of-spheres/107531#107531 Answer by Fernando Muro for Whitehead product with identity on homotopy groups of spheres Fernando Muro 2012-09-19T06:52:10Z 2012-09-19T09:13:07Z <p>As you probably know, the Whitehead product is a degree $-1$ Lie bracket on homotopy groups, i.e. it is graded anticommutative and satisfies the graded Jacobi identity, but not $[x,x]=0$. In particular your maps are homomorphisms. As for the first one, it need not be trivial, if that's what you think. Since suspensions of Whitehead products vanish, your maps are trivial when the target is stable, e.g. for $n\geq i+1$ in the first case. Also, in the first case, it is the kernel of the suspension map in the critical dimension, i.e. $$\pi_n(S^n)\stackrel{Wh_1}\longrightarrow \pi_{2n-1}(S^n)\stackrel{\Sigma}\longrightarrow \pi_{2n}(S^{n+1})$$ is an exact sequence, e.g. for $n=2$ this looks like as follows $$\mathbb{Z}\stackrel{2}\longrightarrow\mathbb{Z}\twoheadrightarrow \mathbb{Z}/2.$$ This uses the Blakers-Massey theorem and the fact that $1\in\pi_n(S^n)$ is a generator. Hence, the analog exact sequence works in the second case only in the special case that $f\in \pi_m(S^n)$ is a generator. Probably more things can be deduced from the elementary properties of primary homotopy operation, but this is what comes to my mind right now.</p> http://mathoverflow.net/questions/134144/functor-of-artinian-rings-in-deformation-theory Comment by Fernando Muro Fernando Muro 2013-06-19T13:12:19Z 2013-06-19T13:12:19Z Maybe if you say what is $\hat G$ http://mathoverflow.net/questions/134063/braided-coverings-and-braided-monodromy Comment by Fernando Muro Fernando Muro 2013-06-18T15:39:46Z 2013-06-18T15:39:46Z I don't understand the question, but it looks like if I should... http://mathoverflow.net/questions/134057/is-there-an-accepted-definition-of-infty-infty-category Comment by Fernando Muro Fernando Muro 2013-06-18T15:36:40Z 2013-06-18T15:36:40Z No. If there were such a consensus, you would have already got some answers. Probably, if you get concrete answers, you will be offered proposals, but no general consensus. http://mathoverflow.net/questions/134057/is-there-an-accepted-definition-of-infty-infty-category Comment by Fernando Muro Fernando Muro 2013-06-18T14:02:29Z 2013-06-18T14:02:29Z what do you mean by 'excepted'? http://mathoverflow.net/questions/133965/p-groups-with-isomorphic-automophism-groups Comment by Fernando Muro Fernando Muro 2013-06-17T16:59:58Z 2013-06-17T16:59:58Z You could edit your question accordingly, adding also the reasons why you think it's true. http://mathoverflow.net/questions/133965/p-groups-with-isomorphic-automophism-groups Comment by Fernando Muro Fernando Muro 2013-06-17T16:27:48Z 2013-06-17T16:27:48Z Are you sure that your assertion is true? http://mathoverflow.net/questions/133881/maximum-likelihood-and-least-squares Comment by Fernando Muro Fernando Muro 2013-06-16T12:19:22Z 2013-06-16T12:19:22Z Take another book. This is clearly not a research question. http://mathoverflow.net/questions/133817/is-there-a-scheme-corresponding-to-the-unit-interval Comment by Fernando Muro Fernando Muro 2013-06-15T16:34:04Z 2013-06-15T16:34:04Z @Martin, you're welcome. http://mathoverflow.net/questions/133817/is-there-a-scheme-corresponding-to-the-unit-interval Comment by Fernando Muro Fernando Muro 2013-06-15T15:09:32Z 2013-06-15T15:09:32Z 'Most' model categories do not have an interval. The idea that motivic homotopies are defined from a product with the affine line may be helpful for some purposes, but also misleading sometimes, as you suggest. Alternatively, think that motivic homotopy theory is defined by forcing the affine line to be contractible. Contractible objects may be used as 'parameter spaces' for homotopies, but while there are many contractible objects in model categories, there is seldom something as good as the interval in topology. http://mathoverflow.net/questions/133816/problem-5-15-i-martin-isaacs-character-theory Comment by Fernando Muro Fernando Muro 2013-06-15T10:04:31Z 2013-06-15T10:04:31Z I do believe that posting an exercise is off-topic in any case. One can, instead, post a problem which coincides with a book's exercise. The difference may only be in the way of posing, not in the actual contents, but it's an important difference, I think. http://mathoverflow.net/questions/133675/a-beck-chevalley-type-condition/133701#133701 Comment by Fernando Muro Fernando Muro 2013-06-14T06:43:25Z 2013-06-14T06:43:25Z You 're welcome! http://mathoverflow.net/questions/133592/equivariant-singular-homology Comment by Fernando Muro Fernando Muro 2013-06-13T23:21:00Z 2013-06-13T23:21:00Z I was guessing. http://mathoverflow.net/questions/133146/homeomorphisms-and-disjoint-unions/133596#133596 Comment by Fernando Muro Fernando Muro 2013-06-13T08:05:48Z 2013-06-13T08:05:48Z Beautiful!!!!!! http://mathoverflow.net/questions/133592/equivariant-singular-homology Comment by Fernando Muro Fernando Muro 2013-06-13T07:50:24Z 2013-06-13T07:50:24Z @Chris, Does Ken Brown's book deal also with Lie groups? http://mathoverflow.net/questions/133566/minimum-value-of-a-trigonometric-equation-is-given-the-problem-is-when-the-minim Comment by Fernando Muro Fernando Muro 2013-06-12T22:13:16Z 2013-06-12T22:13:16Z If it's known and you can't prove it, study other people's proofs.