12
$\begingroup$

Given an extension of $R$-modules $0\to B\to X\to A \to 0$, one usually associates $x\in\operatorname{Ext}^1(A,B)$ to this extension by taking the long exact sequence

$$\dotsb\to \operatorname{Hom}(A,X) \to \operatorname{Hom}(A,A) \xrightarrow{\partial} \operatorname{Ext}^1(A,B)\to \dotsb$$

and setting $x=\partial(\mathrm{id}_A)$. Alternatively one could apply $\operatorname{Ext}^{*}(-,B)$ to get

$$\dotsb\to \operatorname{Hom}(X,B) \to \operatorname{Hom}(B,B) \xrightarrow{\partial} \operatorname{Ext}^1(A,B)\to \dotsb$$

and take $y=\partial(\mathrm{id}_B)$. Do we get the same elements in this way? I.e. is $x=y$? Optimally, can you show this from the standard properties of $\operatorname{Ext}$? I became interested in this because it seems to be necessary to solve a more particular question about a proof I had.

$\endgroup$
4
  • $\begingroup$ Humerously enough I am working on this in my master thesis :) The equivalence extends furhter than just a 1-extension. Unfortunately I do not hold the naswer as of yet. If it's unanswered when I get to it I'll drop an answer! $\endgroup$
    – Zelos Malum
    Mar 7, 2016 at 8:10
  • $\begingroup$ This is related to Weibel Theorem 2.7.6. $\endgroup$
    – tj_
    Mar 12, 2016 at 14:22
  • $\begingroup$ Yeah, that makes sense, I suppose you need to track how the isomorphism given there looks exactly. $\endgroup$
    – Nikolas Kuhn
    Mar 13, 2016 at 23:13
  • $\begingroup$ The connecting homomorphism $\partial$, in both cases, is multiplication by the (class of the) short exact sequence $0 \to B \to X \to A \to 0$ (so long as you think of the ext-groups as classes of $n$-extensions). So you do get the same thing because multiplying on the left or right with identity maps does nothing. I think if you want to get a more precise answer you should say exactly how you want to think of elements of $\operatorname{Ext}^1(A,B)$: classes of SES? of maps on a fixed projective resolution of $A$? something else? $\endgroup$
    – Matthew Towers
    Nov 20, 2017 at 18:48

2 Answers 2

Reset to default
7
$\begingroup$

We can compute this in the derived category.

Extensions give distinguished triangles, and are determined by the corresponding morphism $f : A \to B[1]$.

The two methods you describe for associating an element to the extension correspond to pre- and post-composition with $f$:

$$ \hom(A, A) \xrightarrow{g \mapsto f\circ g} \hom(A, B[1]) $$ $$ \hom(B[1], B[1]) \xrightarrow{h \mapsto h\circ f} \hom(A, B[1]) $$

and so the same element $f \in \hom(A, B[1])$ is indeed obtained by applying these maps to the respective identity morphisms.

$\endgroup$
1
$\begingroup$

Nice observation. Actually, I don't know, if the two agree. Also, this is no answer, but a roadmap to attack the problem.

The problem is that (in the notion of Weibel) $$x = \partial(id_A) \in R^\ast Hom(-,B)(A)$$ while $$y = \partial(id_B) \in R^\ast Hom(A,-)(B)$$ So $x, y$ don't belong to the same set and can't be compared directly.

Let $P \to A$ be a projective resolution and $B \to I$ an injective resolution. Then, as in the proof of Weibel 2.7.6 there are natural isomorphisms
$$H^\ast Hom(A,I) \xrightarrow{f} H^\ast\operatorname{Tot} Hom(P,I) \xleftarrow{g} H^\ast Hom(P,B).$$

So what one could hope for is: $g(x)=f(y)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .