Bijection between Extensions and Ext (Weibel Theorem 3.4.3)

Question

I was wondering about one step in the proof of surjectivity of $\Theta$ constructed for Theorem 3.4.3 in Weibel's "An introduction to homological Algebra".

For an extension $\xi:0\to B\to X\to A\to0$ of $A$ by $B$, he associates the element $x=\Theta(\xi)\in \operatorname{Ext}^1(A,B) $, by applying $\operatorname{Ext}^*(A,-)$ to form the long exact sequence $$\cdots\to\operatorname{Hom}(A,X)\to \operatorname{Hom}(A,A)\xrightarrow{\partial}\operatorname{Ext}^1(A,B)\to\cdots, $$ and setting $x=\partial(\operatorname{id}_A)$. He then shows that $\Theta$ gives a well defined map from the set of equivalence classes of extensions of $A$ by $B$ to $\operatorname{Ext}^1(A,B)$.

In proving surjectivity of $\Theta$, he considers an exact sequence $0\to M\to P \to A\to 0$ with $P$ projective. Applying $\operatorname{Ext}^*(-,B)$ gives $\operatorname{Hom}(M,B)\xrightarrow{\partial}\operatorname{Ext}^1(A,B)\to 0$, so for $x\in \operatorname{Ext}^1(A,B)$ he picks $\beta:M\to B$ with $x=\partial(\beta)$. Then he constructs a diagram

\begin{array}{ccccccccc} 0 & \xrightarrow{} & M & \xrightarrow{} & P & \xrightarrow{} & A & \xrightarrow{} & 0\\ & & \downarrow & & \downarrow & & \parallel & & \\ 0 & \xrightarrow{} & B & \xrightarrow{} & X & \xrightarrow{} & A & \xrightarrow{} & 0 \end{array} with the map from $M$ to $B$ given by $\beta$ and $X$ is the pushout of $B\leftarrow M\to P$. One shows that the lower row is exact (no problem).

Then he claims that the extension given by the lower row maps to $x$ under $\Theta$. How does this follow? He states that one uses the naturality of $\partial$, so one has to apply $\operatorname{Ext}$ in some way.

Edit:Taking the Ext long exact sequence doesn't solve the problem immediately, see the comments below.

Answer 1

The proof as intended by Weibel doesn't seem to work, since it requires one to solve the related question asked here Are those two ways to relate Extensions to Ext equivalent?. However, by using the dual version of his proof we can avoid this issue:

Pick an exact sequence $0\to B \to I\xrightarrow{\pi} N\to 0$, where now $I$ is an injective object. Then apply $Ext(A,-)$ to obtain en exact sequence $$ ... \to Hom(A,N) \xrightarrow{\partial} Ext(A,B) \to 0,$$ and pick $\gamma \in Hom(A,N)$ with $\partial(\gamma)=x$. Now we let $X$ be the pullback of $A\xrightarrow{\gamma} N \xleftarrow{\pi}I$. This fits into a commutative diagram with exact rows:

\begin{array}{ccccccccc} 0 & \xrightarrow{} & B & \xrightarrow{} & X & \xrightarrow{} & A & \xrightarrow{} & 0\\ & & \parallel & & \downarrow & & \downarrow & & \\ 0 & \xrightarrow{} & B & \xrightarrow{} & I & \xrightarrow{} & N & \xrightarrow{} & 0. \end{array} The upper row is now an extension $\xi$ for which one directly sees that $\Theta(\xi)=x$: Applying $Ext(A,-)$ again gives a long ladder diagram, from which we consider the square \begin{array}{ccc} Hom(A,A) & \xrightarrow{\partial'} & Ext^1(A,B) \\ \downarrow& & \parallel \\ Hom(A,N) &\xrightarrow{\partial} & Ext^1(A,B) \end{array} The $\partial$ here is the same as above, and by the definition in Weibel we have $\Theta(\xi) =\partial'(id_A)$. Finally, the left vertical arrow is composition with $\gamma$ by definition of the $Hom$ functor. So $$\Theta(\xi) =\partial'(id_A)=\partial(\gamma\circ id_A) =x.$$

Answer 2

This is a long comment. It doesn't answer the question exactly, but shows how a little modification of the construction described in the question can be used to easily get surjectivity of $\Theta$ without using $\operatorname{Ext}^*(-,B)$. In my opinion usage of $\operatorname{Ext}^*(-,B)$ here just complicates things without benefit.

Let $P_1 \to P_0 \to A$ be the beginning of a projective resolution. Represent $x \in Ext^1(A,B)$ by a cocyle $f: P_1 \to B$. In the same way as indicated in your question you can build a commutative diagramm with exact rows \begin{array}{ccccccccc} & & P_1 & \xrightarrow{d_1} & P_0 & \xrightarrow{\varepsilon} & A & \xrightarrow{} & 0\\ & & f \downarrow & & g \downarrow & & \parallel & & \\ 0 & \xrightarrow{} & B & \xrightarrow{i} & X & \xrightarrow{\rho} & A & \xrightarrow{} & 0 \end{array} Then it follows right from the definition of the connecting homomorphism $\partial: Ext^0(A,A) \to Ext^1(A,B)$ that $x= \partial(id_A)$ and hence the surjectivity of $\Theta$.

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Added: Making the last part precise.

There is a commutative diagramm

\begin{array}{ccccc} & & & & Hom(A,A) \\ & & & & \downarrow \varepsilon^\ast\\ Hom(P_0,B) & \hookrightarrow & Hom(P_0,X) & \overset{\rho^\ast}{\twoheadrightarrow} & Hom(P_0,A) \\ \downarrow & & d_1^\ast\downarrow & & \downarrow \\ Hom(P_1,B) & \overset{i^\ast}{\hookrightarrow} & Hom(P_1,X) & \twoheadrightarrow & Hom(P_1,A) \end{array} with exact rows and exact column on the right.

$id_A$ is represented in $Ext^0(A,A)$ by $\varepsilon^\ast(id_A)=\varepsilon$. The connecting hom. $\partial(\varepsilon)$ is defined as follows:

1. Choose a lift of $\varepsilon$ for $\rho^\ast$: We take $g$.
2. There is a unique $h: P_1 \to B$ with $i^\ast(h)=d_1^\ast(g)$. $\,\partial(\varepsilon)$ is represented by $h$.

By the commutaive diagramm above, $i \circ f = g \circ d_1$. Thus $f = h$. So we have shown: $\partial(id_A) = [f] \overset{\operatorname{def}}{=}x$. qed.

Answer 3

It suffices you undestand how, given a map $\beta: K\to M$ and an extension $x\in{\rm Ext}^1(A,K)$, you can obtain an extension $\beta x=y\in {\rm Ext}^1(A,B)$. The answer is that if the extension $x$ is represented by a short exact sequence $$0\to K\to X\to A\to 0$$ then $\beta x$ is obtained precisely by pushing out by $\beta$, as Weibel did. Similarly, you can show that given $\alpha:A\to A'$, the map induced on ${\rm Ext}^1(A',B)\to {\rm Ext}^1(A,B)$ is obtained by pulling back an extension represented by $$0\to B\to X\to A'\to 0$$ by $\alpha$.