Morphism extends iff image trought connecting map is zero

Question

I am studying the $\operatorname{Ext}$ functor, and I've found the following problem. It is from Assem's Àlgebres et Modules. Assume that all modules are over a ring $R$.

Let $0 \rightarrow M' \rightarrow M \rightarrow M' \rightarrow 0$ be a short exact sequence. Show that a morphism $f': M' \rightarrow N$ extends to a morphism $M \rightarrow N$ if, and only if, the image of $f'$ trought the connecting morphism $\operatorname{Hom}_R(M',N) \rightarrow \operatorname{Ext}_A^1(M'',N)$ is zero.

What I've done: For the $(\Rightarrow)$ implication, since we have the diagram $\require{AMScd}$ \begin{CD} 0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\ @. @V{f'}VV \\ @. N \end{CD} in which the line is exact, we can complete it to an commutative diagram $\require{AMScd}$ \begin{CD} 0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\ @. @V{f'}VV @V{g'}VV @V{1}VV \\ 0 @> >> N @>{u'}>> N' @>{v'}>> M'' @> >> 0 \end{CD} in which both lines are short exact sequences, and $N'$ is the pushout of $(u,f')$. From here, I think that the next step might involve some extensions classes, but I do not know how to proceed.

I was able to do little to nothing with this problem. One of the main issues is that I do not concretely understand the $\operatorname{Ext}$ functor.

If someone could give me some ideas on how to proceed, it would help a lot.

Answer 1

Let $I$ be an injective module containing $N$, and let $Q$ denote the quotient $I/N$. By the injective property of $I$ we may construct a map $f$ (hence also $f''$) such that the following diagram commutes.

$\require{AMScd}$ \begin{CD} 0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\ @. @V{f'}VV @VfVV @Vf''VV \\ 0@> >> N @>{\iota}>> I @>{\epsilon}>> Q @> >> 0 \end{CD}

Now $f'\in \operatorname{Hom}_R(M',N)$, and its image under the connecting morphism is the class represented by $f''$ in $\operatorname{Ext}_R^1(M'',N)$.

This class is trivial if and only if $f''$ factors through $\epsilon$:$$f''=\epsilon h,$$ for some $h\colon M''\to I$.

Given such a map $h$, we have $\epsilon(f-hv)=0$, so $$f-hv=\iota g $$ for some map $g\colon M\to N$.

It follows that $\iota(f'-gu)=0$, so $f'=gu$ as required.

Conversely, if we have a map $g\colon M\to N$ such that $f'=gu$, then $(f-\iota g)u=0$, so $$f-\iota g= hv,$$ for some map $h\colon M''\to I$.

It follows that $(f''-\epsilon h)v=0$, so $f''=\epsilon h$ and $f''$ represents the trivial class in $\operatorname{Ext}_R^1(M'',N)$, as required.

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Here is a brief explanation for why the element of $\operatorname{Ext}_R^1(M'',N)$ represented by the cocycle $f''$ is the same as the one you constructed in your attempt.

Consider the following commutative diagram, where the bottom right square is a pull-back diagram:

$\require{AMScd}$ \begin{CD} 0@> >> M' @>{u}>> M @>{v}>> M'' @> >> 0\\ @. @V{f'}VV @Vf\times vVV @V1_{M''}VV \\ 0@> >> N @>{\iota\times 0}>> I\times_{Q}M'' @>{p_{M''}}>> M'' @> >> 0\\ @. @V{1_N}VV @Vp_IVV @Vf''VV \\ 0@> >> N @>{\iota}>> I @>{\epsilon}>> Q @> >> 0 \end{CD}

In general, given a cocycle we can recover an extension by taking pull-backs as above, so the middle row is the extension represented by $f''$.

It is an easy check that the top left square is a pushout diagram, so the middle row is just your short exact sequence:

\begin{CD} 0 @> >> N @>{u'}>> N' @>{v'}>> M'' @> >> 0 \end{CD}

Answer 2

It may be more helpful to give you an answer along the lines begun in your attempt:

You are correct that the image of $f'$ in $\operatorname{Ext}_R^1(M'',N)$ is represented by your extension:

\begin{CD} 0 @> >> N @>{u'}>> N' @>{v'}>> M'' @> >> 0 \end{CD}

I have added a brief explanation to the bottom of my previous answer, for why this is the same as what I stated in my previous answer.

Now an extension represents the zero element in $\operatorname{Ext}_R^1(M'',N)$ precisely when it splits: In this case when there is a map $s\colon N'\to N$ such that $su'=1_N$.

Suppose such a map $s$ exists. Then $sg'\colon M \to N$ is the required extension of $f'$. Here $g'$ is the map you defined in your commutative diagram.

Conversely given an extension of $f'$ to a morphism $f\colon M \to N$, we may define the splitting $s\colon N'\to N$ by $$s(n,m)=n+f(m).$$