User matthias k&#252;nzer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:28:42Z http://mathoverflow.net/feeds/user/9300 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104941/group-rings-isomorphic-over-f-p-but-not-over-z-p Group rings isomorphic over F_p, but not over Z_p ? Matthias Künzer 2012-08-17T19:51:25Z 2012-08-17T19:51:25Z <p>Suppose given a prime $p$.</p> <p>Question: Do there exist finite groups $G$ and $H$ such that ${\bf F}_p G$ is isomorphic to ${\bf F}_p H$, but such that ${\bf Z}_p G$ is not isomorphic to ${\bf Z}_p H$ ?</p> <p>Variants: Suppose given $s\geqslant 2$ and replace ${\bf Z}_p$ resp. ${\bf F}_p$ by ${\bf Z}/p^s$.</p> <p>Variant: Suppose $G$ and $H$ to be $p$-groups. (It is unknown whether there are nonisomorphic $p$-groups with isomorphic group rings over ${\bf F}_p$ , but still, maybe someone knows an argument in favour of ${\bf F}_p G \simeq {\bf F}_p H$ $\Rightarrow$ ${\bf Z}_p G \simeq {\bf Z}_p H$ in this case?)</p> http://mathoverflow.net/questions/88608/non-finitely-generated-module-is-union-of-countable-chain Non-finitely-generated module is union of countable chain? Matthias Künzer 2012-02-16T07:50:29Z 2012-02-16T08:15:12Z <p>Let $R$ be a ring. Let $M$ be a left $R$-module. </p> <p>Then: $M$ is not finitely generated &lt;=> $M$ is the union of a set of proper submodules closed under binary sums. To recall why: (&lt;=) If $M$ would be f.g., then chosen finitely many generators appear in certain members of the given set of proper submodules. Hence, this set being closed under binary sums, they all appear in one such submodule. So this submodule is not properly contained in $M$, contradiction. (=>) Pick the set of finitely generated submodules of $M$. </p> <p>Strengthening the RHS of this equivalence a bit led me to the <b>question</b>: Can it happen that $M$ is not f.g., but also not the union of an $\mathbb{N}$-indexed chain of proper submodules? I.e. "does (=>) break down if one wants an $\mathbb{N}$-indexed chain"?</p> <p>If $R$ is a field and $M$ an infinite-dimensional vector space, such a chain exists: pick a basis of $M$, write it as $B\sqcup \{ b_i : i \in \mathbb{N} \}$ and form the chain $\langle B\sqcup\{b_1\}\rangle\subset\langle B\sqcup\{b_1,b_2\}\rangle\subset\dots$. </p> <p>But for general $R$, I suspect a different behaviour.</p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/78003#78003 Answer by Matthias Künzer for Awfully sophisticated proof for simple facts Matthias Künzer 2011-10-13T06:42:05Z 2011-10-13T06:42:05Z <p>(1) Let $G$ be a finite group. Let $H\leqslant G$ be a subgroup of index $2$. Let us prove that $H$ is normal in $G$. Let $L|K$ be a Galois extension of fields with Galois group $G$ (easily constructed via a representation of $G$ as a permutation group, taking $L$ to be a function field in suitably many variables on which $G$ acts and $K$ to be the fixed field under $G$). Let $F$ be the fixed field in $L$ under $H$. Then $F|K$ is a quadratic extension, hence normal. By the Main Theorem of Galois Theory, it follows that $H$ is normal in $G$.</p> <p>(2) Let $G$ be a finite group. Let $K$ be a finite field of characteristic not dividing $|G|$. Let us prove Maschke's Theorem in this situation: $KG$ is semisimple. Given two finite dimensional $KG$-modules $X$ and $Y$, it suffices to show that $\text{Ext}^1_{KG}(X,Y) = 0$. But $\text{Ext}^1_{KG}(X,Y) = \text{H}^1(G,\text{Hom}_K(X,Y)) = 0$, since $|G|$ and $|\text{Hom}_K(X,Y)|$ are coprime.</p> <p>(Well, not sure whether any of these arguments are really awfully sophisticated. It's rather breaking a butterfly on a small wheel.)</p> http://mathoverflow.net/questions/62386/examples-of-brown-cofibration-categories-that-are-not-quillen-model-categories Examples of Brown (co)fibration categories that are not Quillen model categories? Matthias Künzer 2011-04-20T08:00:09Z 2011-09-30T11:18:35Z <p>K.S. Brown has shown that much of abstract homotopy theory can be carried out in the setting of Brown (co)fibration categories [MR0341469]. The decisive property, immediate from the axioms, is that any morphism can be factored into a cofibration, followed by a weak equivalence.</p> <p>H.-J. Baues [MR0985099], Cisinski [MR2729017] and Radulescu-Banu [arxiv.org/abs/math/0610009] then followed a similar path. Now someone wants to convince me that this is the proper setting for abstract homotopy theory. To begin with, I do like the simplicity of the axioms. Still, I'd like to be convinced of the practical necessity of this approach. Therefore my question:</p> <p>Are there examples of Brown (co)fibration categories that are not already Quillen model categories?</p> <p>More precisely, does there exist a pair (C,W) consisting of a category C and a subset W in Mor C (weak equivalences) such that (C,W) can be equipped with the structure of a Brown fibration (or cofibration) category, but not with the structure of a Quillen model category?</p> <p>I would particularly be interested in examples in which the lifting axioms of Quillen are the obstacle. If they fail to be a Quillen model category just because they lack limits or colimits, I would be less enthusiastic.</p> <p>On the other hand, I would also welcome examples in which it is relatively easy to show that they are Brown, but relatively hard that they are Quillen. </p> <p>I would also welcome good general stability properties. For instance, in the Brown case there are no big problems if you want to enlarge the set of weak equivalences, as long as the larger set satisfies (2 of 3) and as long as the resulting larger set of acyclic cofibrations is stable under pushouts (incision).</p> <p>To summarise, I want to be able to exclaim: "Good that we have the Brown apparatus!"</p> http://mathoverflow.net/questions/74985/number-of-subgroups-of-p-groups/75005#75005 Answer by Matthias Künzer for Number of Subgroups of p-Groups Matthias Künzer 2011-09-09T15:07:23Z 2011-09-09T15:07:23Z <p>Wrong for groups of order $3^4$ or $5^4$.</p> <p>Here is Magma code (can be tested in <a href="http://magma.maths.usyd.edu.au/calc/" rel="nofollow">http://magma.maths.usyd.edu.au/calc/</a> ):</p> <p>p := 5;</p> <p>a := 5;</p> <p>ord := p^a;</p> <p>n := NumberOfSmallGroups(ord);</p> <p>for i in [1..n] do </p> <p>G := SmallGroup(ord,i);</p> <p>ST := [SubgroupLattice(G)[i] : i in [1..#SubgroupLattice(G)]];</p> <p>print [&amp;+[Integers()!(Order(G)/Order(Normaliser(G,H))) : H in ST | IsIsomorphic(H,K)] mod p : K in ST];</p> <p>end for;</p> <p>Testing groups of order $5^5$, we see that the number in question is also not necessarily congruent to -1, 0 or +1 modulo p.</p> http://mathoverflow.net/questions/67335/heller-operator-without-left-adjoint Heller operator without left adjoint? Matthias Künzer 2011-06-09T08:47:21Z 2011-06-09T08:47:21Z <p>Suppose given a noetherian ring $R$. On the stable category $R\text{-}\underline{\text{mod}} := R\text{-mod}/R\text{-proj}$, we have the Heller operator $$\Omega : R\text{-}\underline{\text{mod}} \to R\text{-}\underline{\text{mod}} \; ,$$ defined by a choice of short exact sequences $\Omega X \to P \to X$ with $P$ projective, accordingly on morphisms. </p> <p><b>Does there exist a ring $R$ for which $\Omega$ does not have a left adjoint?</b></p> <p>Of course, $R$ should neither be self-injective ($\Omega$ an equivalence) nor hereditary ($\Omega = 0$).</p> <p>If $\Omega$ has a left adjoint $S$, then $(X,\Omega f) = (SX,f)$ is injective for $X$ an $R$-module and $f$ a monomorphism. So I've been searching (without success) for a monomorphism $f$ in the stable category that is not mapped to a monomorphism under $\Omega$. First of all, $f$ should be a non-split mono, but this is possible, since we are not in the triangulated case. </p> <p>It was pointed out to me that in "On a theorem of E. Green on the dual of the transpose" by M. Auslander and I. Reiten, it is shown that that Omega lifts to an exact functor isomorphic to tensoring with a bimodule if $R$ is an artin algebra (MR1143845).</p> <p>The question is related to <a href="http://mathoverflow.net/questions/62386/examples-of-brown-cofibration-categories-that-are-not-quillen-model-categories" rel="nofollow">this question</a> ("Brown, but not Quillen?").</p> http://mathoverflow.net/questions/62218/german-mathematical-terms-like-nullstellensatz/62376#62376 Answer by Matthias Künzer for German mathematical terms like "Nullstellensatz" Matthias Künzer 2011-04-20T06:08:50Z 2011-04-20T06:08:50Z <p>The practice to use Gothic letters sometimes for ideals ($\mathfrak{a}$, $\mathfrak{b}$, ...) and often for Lie algebras ($\mathfrak{g}$, $\mathfrak{h}$, ..) seems to be of German origin.</p> <p>Also to use the lesser known "kernel" instead of the better known "core" seems to stem from the German "Kern".</p> http://mathoverflow.net/questions/59265/pi-nf0-implies-f-simeq-ast-h-nf0-implies-f-simeq-ast/59268#59268 Answer by Matthias Künzer for $\pi_n(f)=0$ implies $f\simeq \ast$? $H_n(f)=0$ implies $f\simeq \ast$? Matthias Künzer 2011-03-23T08:20:11Z 2011-03-23T08:20:11Z <p>Via Dold-Kan, the $\pi_n$-question in the particular case of simplicial abelian groups can be reformulated in the category of complexes of abelian groups: Given a morphism $f : X\to Y$ of bounded below complexes of abelian groups such that $\text{H}_\ast f = 0$ for all $n$, is $f = 0$ in the derived category? This is not the case in general.</p> <p>Take $A$ and $B$ two abelian groups such that $\text{Ext}^1(A,B)\ne 0$, e.g. $A = B = \mathbf{Z}/2$. Let $P$ be a projective resolution of $A$. Let $Q$ be a projective resolution of $B$. Then Hom_{K(Z)}(P,Q[1]) = Hom_{D(Z)}(P,Q[1]) = Hom_{D(Z)}(A,B[1]) = Ext^1(A,B) =/= 0 (TeX didn't work). Let $f : P\to Q[1]$ be a morphism of complexes that is not homotopic to zero. Then it is nonzero in the derived category. But $\text{H}_n f = 0$ for all $n$, since $\text{H}_n(P) = 0$ if $n\ne 0$ and $\text{H}_n(Q[1]) = 0$ if $n\ne 1$. </p> http://mathoverflow.net/questions/57904/why-not-define-triangulated-categories-using-a-mapping-cone-functor/57937#57937 Answer by Matthias Künzer for Why not define triangulated categories using a mapping cone functor? Matthias Künzer 2011-03-09T09:59:01Z 2011-03-09T09:59:01Z <p>Verdier, Astérisque 239, Ch. II, Prop. 1.2.13 (p. 104) says that a triangulated category (with countable coproducts or products) equipped with a cone functor has to split. </p> <p>Ben Wieland is right - you get a cone functor when working with dérivateurs (or filtered derived categories), but that functor is no longer intrinsic to the base category, and you have to carry diagram categories along. </p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/49508#49508 Answer by Matthias Künzer for Spectral sequences: opening the black box slowly with an example Matthias Künzer 2010-12-15T10:14:50Z 2010-12-15T10:14:50Z <p>Verdier and Deligne introduced more terms in a spectral sequence of a filtered complex (indexed by 4+1 indices instead of 2+1), thus in particular factoring the differentials into epis and monos. This enables one to splice certain short exact sequences to obtain Massey triangles or to obtain H(E_r) = E_{r+1}. Not sure whether this helps for calculations, but it does help with understanding what's going on with differentials. Cf. Deligne, Décompositions dans la catégorie dérivée, appendix, MR1265526; cf. also Verdier, Des catégories dérivées des catégories abéliennes, MR1453167.</p> http://mathoverflow.net/questions/46877/natural-examples-of-sequences-of-adjoint-functors/47614#47614 Answer by Matthias Künzer for Natural examples of sequences of adjoint functors Matthias Künzer 2010-11-28T22:01:51Z 2010-11-28T22:01:51Z <p>Similar to Ben Wieland's and Sasha's answer: Let $\mathcal{C}$ be the category of complexes in an abelian category. Let $\underline{\mathcal{C}(\Delta_0)}$ be the homotopy category of $\mathcal{C}$. Let $\underline{\mathcal{C}(\Delta_1)}$ be the category of arrows with values in $\mathcal{C}$, with pointwise homotopy equivalences formally inverted (not to be confused with $\underline{\mathcal{C}(\Delta_0)}(\Delta_1)$). The functor $\underline{\mathcal{C}(\Delta_0)}\to\underline{\mathcal{C}(\Delta_1)}$ that sends an object of $\mathcal{C}$ to the identity arrow on this object, morphisms accordingly, gives rise to an infinite chain of adjoint functors ("walking along a distinguished triangle"). [Which functors, such as $\Delta_1\to\Delta_0$ considered here, have this property?]</p> http://mathoverflow.net/questions/39431/the-residue-class-functor-from-a-frobenius-category-to-its-stable-category-induce The residue class functor from a Frobenius category to its stable category induces a functor on cube-shaped diagrams - is it dense? Matthias Künzer 2010-09-20T20:31:42Z 2010-09-21T09:42:58Z <p>Let $\mathcal{E}$ be a Frobenius category, i.e. an exact category with sufficiently many bijective objects. (Such as e.g. the category of complexes over an additive category.)</p> <p>Let $\underline{\mathcal{E}}$ be its stable category, defined as the factor category of $\mathcal{E}$ modulo its full subcategory of bijective objects. We have the residue class functor $R:\mathcal{E}\to\underline{\mathcal{E}}$. (For complexes, this is the canonical functor from the category of complexes to its homotopy category.)</p> <p>Consider the category $D := \Delta_1\times\Delta_1\times\Delta_1$, i.e. a cube. Denote by $\mathcal{E}(D)$ the category of functors from $D$ to $\mathcal{E}$, i.e. the category of $\mathcal{E}$-valued diagrams of shape $D$. Etc.</p> <p>Consider the functor $R(D):\mathcal{E}(D)\to\underline{\mathcal{E}}(D)$, obtained by pointwise application of $R$. Is this functor $R(D)$ dense, i.e. surjective on isoclasses? In other words, given a stably commutative cube-shaped diagram, does there exist a strictly commutative cube-shaped diagram isomorphic to it in $\underline{\mathcal{E}}(D)$?</p> <p>My guess would be: no. But I was unable to construct a counterexample. I was searching in $\mathcal{E} = {\bf Z}/p^a\text{-mod}$.</p> <p>The paper [1] deals with the question in greater generality, but it doesn't seem to give an explicit answer to my question. (This question can be generalised: one does not have to stick to the Frobenius case, and one can ask for obstructions depending on $D$ against the density of $R(D)$.)</p> <p>[1] Dwyer, W. G.; Kan, D. M.; Smith, J. H., Homotopy commutative diagrams and their realizations, J. Pure Appl. Alg. 57, p. 5-24, 1989.</p> http://mathoverflow.net/questions/4653/classifying-triangulated-structures-on-a-graded-category/39245#39245 Answer by Matthias Künzer for Classifying triangulated structures on a graded category Matthias Künzer 2010-09-18T20:45:45Z 2010-09-18T20:45:45Z <p>Heller constructs a bijection between (Puppe) triangulations on a triangulable category C (with given shift functor) and certain isomorphisms between certain shift functors in Theorem 16.4 of "Stable Homotopy Categories", Bull. Am. Math. Soc. 74, p. 28-63, 1968. He concludes in Corollary 16.5 that a subgroup of the automorphism group of the identity functor of the stable category of the Freyd category of C acts freely and transitively on the set of triangulations on C (with given shift functor). </p> <p>In §17 he uses this to show that there are uncountably many triangulations on C in the case C := stable homotopy category of finite CW-complexes (cf. p. 47, l. 19).</p> <p>Not sure about your extra condition ("enriched in graded abelian groups"). The shift functor is the same for all triangulations exhibited by Heller, though.</p> http://mathoverflow.net/questions/32755/is-there-any-generalization-of-the-dold-kan-correspondence/39090#39090 Answer by Matthias Künzer for Is there any generalization of the Dold-Kan correspondence? Matthias Künzer 2010-09-17T12:30:07Z 2010-09-17T12:30:07Z <p>For simplicial groups, there is the generalised Dold-Kan-Puppe theorem by Carrasco-Cegarra: "Group-theoretic algebraic models for homotopy types", J. Pure Appl. Alg. 75 (1991), no. 3, 195--235 (see also MathSciNet MR1137837). (The additive notation in the article is to be read non-commutatively.)</p> http://mathoverflow.net/questions/33345/if-a-colimit-of-distinguished-triangles-exists-is-it-also-a-distinguished-triang/39086#39086 Answer by Matthias Künzer for If a colimit of distinguished triangles exists, is it also a distinguished triangle? Matthias Künzer 2010-09-17T11:05:08Z 2010-09-17T11:05:08Z <p>(Deligne has an assertion that a certain triangle in the ind-category of a triangulated category can be written as a limit of distinguished triangles there; see SGA 4, XVII, Lemme 1.2.2.1, <a href="http://www.msri.org/publications/books/sga/sga/4-3/4-3t_275.html" rel="nofollow">http://www.msri.org/publications/books/sga/sga/4-3/4-3t_275.html</a> . Does not seem to answer the question, though.)</p> http://mathoverflow.net/questions/24524/can-one-extend-a-morphism-of-commutative-triangles-to-a-morphism-of-octahedral-d/39080#39080 Answer by Matthias Künzer for Can one extend a morphism of commutative triangles to a morphism of octahedral diagrams? Matthias Künzer 2010-09-17T10:29:28Z 2010-09-17T10:29:28Z <p>The identity of commutative triangles does not lift to a morphism of Verdier octahedra in general, for it may happen that there exist two mutually nonisomorphic Verdier octahedra on the same commutative triangle. Cf. <a href="http://www.math.rwth-aachen.de/~kuenzer/counterexample.pdf" rel="nofollow">http://www.math.rwth-aachen.de/~kuenzer/counterexample.pdf</a> . This has been observed already by Amnon Neeman in the 90s (unpublished). </p> <p>(The situation is better if one uses 3-triangles, i.e. "distinguished octahedra", in the sense of Heller/oo-triangulated categories. Cf. <a href="http://www.math.rwth-aachen.de/~kuenzer/heller.pdf" rel="nofollow">http://www.math.rwth-aachen.de/~kuenzer/heller.pdf</a> , Lem. 3.2. So there exists a choice of octahedra such that morphisms of commutative triangles lift to morphisms of these particular octahedra.)</p> http://mathoverflow.net/questions/114553/loop-space-of-a-category Comment by Matthias Künzer Matthias Künzer 2012-12-26T10:24:02Z 2012-12-26T10:24:02Z In Sec. 4 of Ch. VII, he constructs the path category _Ch_(X). In Sec. 12, he uses an alternative construction _Ch_oo(X) to produce an &quot;analogue of the Cartan-Serre formalism&quot;. In Sec. 13, he constructs an equivalence of _Ch_(X) and _Ch_oo(X), saying the former is better. On page 106, then the (co)cone of a morphism appears. Of course, it's still in the form of a &quot;mathematical diary&quot;, but the constructions are there. http://mathoverflow.net/questions/114553/loop-space-of-a-category Comment by Matthias Künzer Matthias Künzer 2012-12-24T10:32:09Z 2012-12-24T10:32:09Z Grothendieck has worked on cat&#233;gories de chemins, see www.math.jussieu.fr/~maltsin/groth/Derivateursengl.html . http://mathoverflow.net/questions/91785/derived-functors-versus-spectral-sequences/91798#91798 Comment by Matthias Künzer Matthias Künzer 2012-08-23T16:41:19Z 2012-08-23T16:41:19Z I think that in &#167;3, the third curly Hom should be over X. http://mathoverflow.net/questions/104941/group-rings-isomorphic-over-f-p-but-not-over-z-p Comment by Matthias Künzer Matthias Künzer 2012-08-20T09:31:05Z 2012-08-20T09:31:05Z @Florian: Thanks! I wouldn't want to make a bet, though. http://mathoverflow.net/questions/104941/group-rings-isomorphic-over-f-p-but-not-over-z-p Comment by Matthias Künzer Matthias Künzer 2012-08-17T20:14:22Z 2012-08-17T20:14:22Z @Florian: Thanks, I've overlooked that. @Chris Gerig: Yes, ${\bf Z}_p$ is just short for $\hat{{\bf Z}_{(p)}}$. http://mathoverflow.net/questions/99506/blackbox-theorems/99557#99557 Comment by Matthias Künzer Matthias Künzer 2012-06-14T15:58:20Z 2012-06-14T15:58:20Z Plus the Gabriel-Quillen-Laumon theorem, embedding an exact category into an abelian one. http://mathoverflow.net/questions/95229/when-is-the-kth-homology-group-of-a-space-isomorphic-to-its-kth-homotopy-group Comment by Matthias Künzer Matthias Künzer 2012-04-26T16:51:55Z 2012-04-26T16:51:55Z Given a dimensionwise free simplicial group G (e.g. obtained when Kan's loop group functor is applied to a reduced simplicial set), we may form the dimensionwise abelianisation G/G'. IIRC the homology of G is the homotopy of G/G', and the Hurewicz map from homotopy to homology fits into a long exact homotopy sequence on G' -&gt; G -&gt; G/G' (MR0098371). http://mathoverflow.net/questions/82087/grothendiecks-manuscript-on-topology Comment by Matthias Künzer Matthias Künzer 2012-03-25T14:35:39Z 2012-03-25T14:35:39Z No, I don't have more than what I've typed. http://mathoverflow.net/questions/82087/grothendiecks-manuscript-on-topology Comment by Matthias Künzer Matthias Künzer 2012-03-25T14:07:39Z 2012-03-25T14:07:39Z I just had a short look on his &quot;topology of forms&quot; manuscripts, someone showed me several bunches of paper. I remember asking whether Grothendieck still deals with topological spaces in the classical sense therein, but I don't remember the answer. Other than that, Pursuing Stacks, D&#233;rivateurs, Longue Marche. http://mathoverflow.net/questions/82087/grothendiecks-manuscript-on-topology Comment by Matthias Künzer Matthias Künzer 2012-03-25T11:34:11Z 2012-03-25T11:34:11Z I would assume that Scharlau believed Grothendieck, who pronounced all of his own manuscripts to be unreadable for anyone but himself. This has turned out not to be true, however. (In the meanwhile, Scharlau's biographies are available via Amazon btw, Parts I and III in German, Part I also in English.) http://mathoverflow.net/questions/89759/cohomological-characterization-of-homotopy-pullbacks Comment by Matthias Künzer Matthias Künzer 2012-03-11T09:23:42Z 2012-03-11T09:23:42Z In K(Z-mod) (so not directly related to your question), Alberto Canonaco has found a commutative quadrangle with isomorphic cones that is not homotopy cartesian. Cf. Canonaco, K., &quot;A sufficient criterion for homotopy cartesianess&quot;. http://mathoverflow.net/questions/88608/non-finitely-generated-module-is-union-of-countable-chain Comment by Matthias Künzer Matthias Künzer 2012-02-17T17:18:36Z 2012-02-17T17:18:36Z Thanks! (I didn't look into that discussion when it was new, and it didn't show up automatically when I posted this question.) It is exactly what I was after. http://mathoverflow.net/questions/88608/non-finitely-generated-module-is-union-of-countable-chain/88610#88610 Comment by Matthias Künzer Matthias Künzer 2012-02-17T17:13:45Z 2012-02-17T17:13:45Z Thanks! Looks already pretty big, compared with what I had tried so far. http://mathoverflow.net/questions/81721/a-homotopy-commutative-diagram-that-cannot-be-strictified Comment by Matthias Künzer Matthias Künzer 2011-12-14T07:10:21Z 2011-12-14T07:10:21Z On the other hand, for which categories I is Top^I -&gt; Ho(Top)^I dense? I.e. what is the condition on I that ensures that there is never an obstruction against rectifying a diagram of shape I? Cf. also <a href="http://mathoverflow.net/questions/39431/the-residue-class-functor-from-a-frobenius-category-to-its-stable-category-induce" rel="nofollow" title="the residue class functor from a frobenius category to its stable category induce">mathoverflow.net/questions/39431/&hellip;</a> . http://mathoverflow.net/questions/80926/when-does-the-forgetful-functor-s-mod-r-mod-induce-injective-maps-on-ext-group/81556#81556 Comment by Matthias Künzer Matthias Künzer 2011-11-22T07:20:30Z 2011-11-22T07:20:30Z Do you need that $M$ is an $S$-module, which becomes an $R$-module after restricting along $R\to S$? If so, isn't $\Omega\otimes_S M \to S\otimes_R M\to M$ in your context split as well, as a sequence of left $S$-modules? So that $\text{Ext}^2_S(-,N)$, being an additive functor, maps it to a split short exact sequence $\text{Ext}^2_S(\Omega\otimes_S M,N)\leftarrow\text{Ext}^2_S(S\otimes_R M,N)\leftarrow\text{Ext}^2_S(M,N)$, which amounts to the split short exact sequence $\text{Ext}^2_S(\Omega\otimes_S M,N)\leftarrow\text{Ext}^2_R(M,N)\leftarrow\text{Ext}^2_S(M,N)$?