User paul vankoughnett - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T16:40:33Z http://mathoverflow.net/feeds/user/27153 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131010/composition-in-the-category-quotient/131023#131023 Answer by Paul VanKoughnett for Composition in the category quotient Paul VanKoughnett 2013-05-18T04:57:24Z 2013-05-18T04:57:24Z <p>First, you actually want $((N'' + N')/N')$ every time you have $(N'' + N')/N$. If this was the source of your confusion, sweet. If not, here goes. (Also, the double underscores in the NUMDAM url got messed up by MO's processing; <a href="http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1962__90_/BSMF_1962__90__323_0/BSMF_1962__90__323_0.pdf" rel="nofollow">here's a working link</a>.)</p> <ol> <li><p>There's a short exact sequence $0 \to M'/M'' \to M/M'' \to M/M' \to 0$, and $M'/M'' \cong N/(N'' + N')$, since $M''$ is the kernel of the composition $M' \to N/N' \to N/(N'' + N')$; finally, $N/(N'' + N')$ is a quotient of $N/N''$, which is in $\mathcal{C}$, so it's in $\mathcal{C}$ as well.</p></li> <li><p>$\mathcal{C}$ is closed under taking subobjects and quotient objects. $N'$ is in $\mathcal{C}$, $N' \cap N''$ is a subobject, and $g(N' \cap N'')$ is a quotient of that.</p></li> <li><p>A sum of two subobjects of another object is a quotient of their <em>direct</em> sum. Here, $P'$ and $g(N'' \cap N')$ are in $\mathcal{C}$, so their direct sum is as well, and so this internal sum is as well.</p></li> <li><p>The point of this was to compose $\overline{f}$ and $\overline{g}$ in the quotient category, but to do this, we had to lift them to maps $f$ and $g$ in the original category. Gabriel's saying that we get the same composition in the quotient category even if we had picked different lifts of $\overline{f}$ and $\overline{g}$. Let's say we picked $\tilde{f}:\widetilde{M'} \to N/\widetilde{N'}$ and $\tilde{g}:\widetilde{N''} \to P/\widetilde{P'}$ instead. Since these have to become $\overline{f}$ and $\overline{g}$ in the direct limit, then, for example, $\widetilde{f}$ and $f$ must be the same as maps from some subobject of $M' \cap \widetilde{M'}$ to some quotient of $N/(N' + \widetilde{N'})$, and likewise for $g$ and $\widetilde{g}$.</p> <p>So we reduce to the case where $\widetilde{f}$ is defined 'further along in the direct system' than $f$. That is, $\widetilde{f}$ is a map $\widetilde{M'} \to N/\widetilde{N'}$ where $\widetilde{M'}$ and $\widetilde{N'}$ are subobjects of $M'$ and $N'$ respectively, with the obvious restrictions on their quotients, and we have to show that $gf$ and $g\widetilde{f}$ become the same map in the quotient category (and then do the same but vary $g$). I'm not going to do this whole proof for you, but if you're confused about things like this, it's probably worth doing yourself. Good luck!</p></li> </ol> http://mathoverflow.net/questions/129837/definition-of-subobject-classifier-in-presheaves/129846#129846 Answer by Paul VanKoughnett for Definition of subobject classifier in presheaves Paul VanKoughnett 2013-05-06T14:28:20Z 2013-05-06T14:28:20Z <p>What's going on is that, since $E$ is a contravariant functor to $\mathbf{Sets}$, by definition $f:D \to C$ induces a map of sets $f^*:E(C) \to E(D)$. Then $u_C(e)$ consists of the arrows $f$ such that $f^*(e)$ actually lands in the subobject $U(D)$ of $E(D)$. You can then check that this is a sieve for each $e$, and so on.</p> <p>It might be helpful to let $E$ be a representable functor, say $E = \mathbf{C}(-,X)$. Then $e$ is a map $C \to X$, $U(D)$ is a set of maps $D \to X$, and $u_C(e)$ is the set of maps $D \to C$ such that the composition $D \to C \stackrel{e}{\to} X$ is in the set $U(D)$.</p> http://mathoverflow.net/questions/123559/injectivity-is-a-local-property-over-noetherian-rings/123570#123570 Answer by Paul VanKoughnett for injectivity is a local property over noetherian rings Paul VanKoughnett 2013-03-04T20:56:02Z 2013-03-04T20:56:02Z <p>Let $A \to B$ be an injection of $R$-modules. We want to show that $\mathrm{Hom}(B,M) \to \mathrm{Hom}(A,M)$ is surjective. It suffices to check this locally (since localization is an exact functor and the cokernel of the map is zero iff it's zero locally). So let $p$ be a prime, and consider the localized map $\mathrm{Hom}_R(B,M) \otimes R_p \to \mathrm{Hom}_R(A,M) \otimes R_p$. Now, for $B$ and $A$ finitely presented over $R$, this is $\mathrm{Hom}_{R_p}(B_p,M_p) \to \mathrm{Hom}_{R_p}(A_p,M_p)$. Again, localization is exact so $A_p \to B_p$ is injective, and $M_p$ is injective by hypothesis, so this map is surjective.</p> <p>What's left is to show that we only need to check injectivity against maps of finitely presented $R$-modules; since $R$ is noetherian, 'finitely presented' is equivalent to 'finitely generated.' In fact, there's a thing called <a href="http://en.wikipedia.org/wiki/Injective_module#Baer.27s_criterion" rel="nofollow">Baer's injectivity criterion</a> that says you only have to check injectivity against inclusions $I \to R$ for $I$ an ideal! So we're done.</p> http://mathoverflow.net/questions/111862/algebraic-closure-as-a-fibrant-replacement Algebraic closure as a fibrant replacement? Paul VanKoughnett 2012-11-09T01:51:25Z 2012-11-09T01:51:25Z <p>Emil Artin's construction of the algebraic closure of a field $K$ is as follows. Let $K_0 = K$, and inductively let <code>$\{x_f\}$</code> be a set of indeterminates indexed by the irreducible $f$ in one variable over $K_i$. Let $\mathfrak{m}_i$ be a maximal ideal of $K_i[x_f]$ containing the ideal generated by all $f(x_f)$ for $f$ irreducible (one can prove this ideal is proper). Define <code>$K_{i+1} = K_i[x_f]/\mathfrak{m}_i$</code>. Clearly, each $K_{i+1}$ is algebraic over $K_i$ (and thus over $K_0$), and every polynomial with coefficients in $K_i$ splits in $K_{i+1}$. In particular, the union $\bar{K}$ of the $K_i$ is algebraic over $K$ and algebraically closed.</p> <p>To me, a fan of homotopy theory, this seems an awful lot like a small object argument. We can express a solution of the polynomial $f$ over the field $K$ as a solution to a lifting problem: we want the map $0:K[x] \to K$ sending $x$ to $0$ to factor through the map $f:K[x] \to K[x]$ sending $x$ to $f(x)$, so we want a lift to a square which has $K \to 0$ on the right, $f:K[x] \to K[x]$ on the left, and $0$ on the top. (I'd draw this here, but I don't know how.) To construct $K_1$, we are doing something like taking the pushout of the diagram $$\bigotimes_f K[x_f] \stackrel{\otimes f}{\leftarrow} \bigotimes_f K[x_f] \stackrel{0}{\rightarrow} K$$ whose left arrow is a coproduct of the left sides of the aforementioned lifting problems. The algebraic closure itself is then the colimit of these pushouts, which is doing something like giving a fibrant replacement for $K$.</p> <p>As you've probably noticed, this naïve outline has quite a few problems. The pushouts constructed above are not actually the fields $K_i$, but rather become them after a quotient by a maximal ideal. Also, the lifting problems we are solving differ between the $K_i$: the left sides of the squares that appear are of the form $f:K_i[x] \to K_i[x]$, where $f$ is an irreducible polynomial over $K_i$. On the other hand, the small object argument is done via a single set of generating cofibrations which appear on the left sides of squares at every stage. </p> <p>I'd like to think that, after passing from the category of $K$-algebras to something presumably more complicated, one can construct a model structure in which the algebraic closure of a field appears as its fibrant replacement. Does anyone know if this is true?</p> http://mathoverflow.net/questions/131010/composition-in-the-category-quotient/131023#131023 Comment by Paul VanKoughnett Paul VanKoughnett 2013-05-19T23:08:38Z 2013-05-19T23:08:38Z $g$ is a map $N'' \to P/P'$; taking a further quotient induces $N'' \to P/P''$; and then by definition, $N' \cap N''$ is a subobject of the kernel of this map, so it passes to $N''/(N'\cap N'')\to P/P''$. There's a good reason why arguments about abelian categories look like arguments about modules: the Freyd-Mitchell embedding theorem, which says that any small abelian category embeds into some category of modules over a ring. So any statement you can phrase in terms of an arbitrary small set of objects in an abelian category, and prove when those objects are modules, is true in general! http://mathoverflow.net/questions/130836/computation-of-extzn-z Comment by Paul VanKoughnett Paul VanKoughnett 2013-05-16T17:27:16Z 2013-05-16T17:27:16Z This should answer your question: <a href="http://mathoverflow.net/questions/40499/extensions-of-an-infinite-product-of-copies-of-z-by-z" rel="nofollow" title="extensions of an infinite product of copies of z by z">mathoverflow.net/questions/40499/&hellip;</a> http://mathoverflow.net/questions/129504/split-and-pure-exact-sequence-of-sheaves Comment by Paul VanKoughnett Paul VanKoughnett 2013-05-03T15:24:32Z 2013-05-03T15:24:32Z Tensor products commute with direct sums, and if the sequence is split, $B \cong A \oplus C$. http://mathoverflow.net/questions/129002/is-there-a-local-projective-model-structure-on-simplicial-sheaves-what-are-its Comment by Paul VanKoughnett Paul VanKoughnett 2013-04-28T23:35:04Z 2013-04-28T23:35:04Z This paper (math.uiuc.edu/K-theory/0462/combination2.pdf) by Benjamin Blander seems to imply that the answer to your first question is 'yes,' and you again get a Quillen equivalent model category, with sheafifying and forgetting being a Quillen equivalence. http://mathoverflow.net/questions/125225/intuition-as-to-why-the-k-theory-of-a-ring-should-be-the-homotopy-theory-of-an-h Comment by Paul VanKoughnett Paul VanKoughnett 2013-03-22T02:34:23Z 2013-03-22T02:34:23Z That isn't really an answer to your questions, so for now let me just make an insinuation: what could you mean by a space 'naturally' having a multiplication on its homotopy groups, other than this being induced by a multiplication on the space itself? http://mathoverflow.net/questions/125225/intuition-as-to-why-the-k-theory-of-a-ring-should-be-the-homotopy-theory-of-an-h Comment by Paul VanKoughnett Paul VanKoughnett 2013-03-22T02:32:28Z 2013-03-22T02:32:28Z In fact the $K$-groups of a ring are naturally modelled as the homotopy groups of a <i>spectrum</i>. (If you want to stick with spaces, you can in turn model this as an infinite loop space, i.e. a space which can be expressed as the form $\Omega^nX_n$ for some space $X_n$, for every $n$.) When you're starting with a ring, the spectrum you get is always a ring spectrum, which is the spectrum version of an $H$-space. I don't actually know why this is true, but I'd guess it pops out of one of the various constructions of this spectrum. http://mathoverflow.net/questions/123501/pi-cohomology-class-a-variant-of-cohomology-class Comment by Paul VanKoughnett Paul VanKoughnett 2013-03-04T03:32:34Z 2013-03-04T03:32:34Z In order for this to be well-defined, you'd need to say precisely what it means for two cycles to '&quot;deform&quot; into each other continuously.' If both cycles are subcomplexes of $X$, you could ask for them to be simplicially homotopic, but then you'd want your simplicial set to be Kan, i. e. probably the singular complex of the space $X$ rather than a triangulation. Then you'd need to make sure that your definition extends well to the abelian group of cycles.