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.

  • $\begingroup$ Have you tried using the long exact ladder associated to that commutative diagram of short exact sequences? $\endgroup$
    – Pedro
    Mar 4, 2016 at 1:00
  • $\begingroup$ Yup, what I struggle with is that when you apply Ext(A,-) to compute $\Theta$, I don't know how to handle the arising map Ext(A,\beta). On the other hand if you apply Ext(-,B) to the diagram you end up with a connecting homomorphism $\partial:\operatorname{Hom}(B,B)\to \operatorname{Ext}(A,B)$, and it is fairly easy to show that it maps id_B to x. However, I cannot show that this gives the same class of Ext(A,B) as using Ext(A,-) and taking $\partial(id_A)$. $\endgroup$ Mar 4, 2016 at 7:44
  • $\begingroup$ More precisely, with the first application of $\operatorname{Ext}$ I can show that $\Theta$ of the extension is $\operatorname{Ext}(A,\beta)\circ \partial (id_A)$. Wit the second that $\partial\circ \beta (id_B) = x$. But I can't fit those two pieces of data together. $\endgroup$ Mar 4, 2016 at 10:47

3 Answers 3


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 $$\Theta(\xi) =\partial'(id_A)=\partial(\gamma\circ id_A) =x.$$

  • $\begingroup$ I think it's OK. There is another downside with Weibel's proof by using injectives as well as projectives: There are categories that have enough injectives but not enough projectives. In this case Weibel's proof doesn't apply at all, while the statement "Ext (defined via injectives) and Extensions are isomorphic" holds over all categories with enough injectives. [same if we interchange injective / projective] $\endgroup$
    – tj_
    Mar 14, 2016 at 3:26

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 $\Theta$.

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.

  • $\begingroup$ Thanks for the idea, but I think this doesn't quite pan out either for basically the same reason: You somehow have to relate the injective resolutions needed to construct the connecting homomorphism to your projective one and you still have to show the problem asked here: math.stackexchange.com/q/1684166/112739 $\endgroup$ Mar 13, 2016 at 23:20
  • $\begingroup$ No, you don't. The purpose of my comment was just to point out that Ext(-,B) is not needed to show the equivalence of Ext and Extensions. Where do you feel the need for using an injective resolution in my comment ? $\endgroup$
    – tj_
    Mar 13, 2016 at 23:55
  • $\begingroup$ Basically, I don't see how "It follows right from the definition of the connecting homomorphism". Can you make this precise? (I'm sorry if I am overseeing something really obvious here). It is also not too hard to see, that the proof Weibel intends works if and only if the two ways to construct $\Theta$ are equivalent. So you would implicitly prove that, which makes me sceptical. $\endgroup$ Mar 13, 2016 at 23:58

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 $\alpha$.

  • $\begingroup$ I don't see how that helps. Is what you are describing not just another way to express the naturality of the connecting homomorphism? $\endgroup$ Mar 4, 2016 at 21:54
  • $\begingroup$ @NikolasKuhn No. I'm telling you how to obtain a representative of $\beta(x)$ given one of $x$, that is, how ${\rm Ext}(A,\beta)$ acts on extensions. $\endgroup$
    – Pedro
    Mar 4, 2016 at 22:25
  • $\begingroup$ Ok, if I understand the first statement in your answer correctly, it is just a consequence of the naturality in the long exact sequence. As for the second statement, I don't see how to show it for much the same reason I struggle with my original problem. If we knew that one could compute $\Theta$ by applying $\operatorname{Ext}(-,B)$ to the extension as in my first comment above, both would follow easily. But that is described nowhere in the book and I don't feel like it follows from the usual properties of $\operatorname{Ext}$. $\endgroup$ Mar 5, 2016 at 8:45

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