Bijection between $\mathrm{Ext}^1$ and equivalence classes of extensions

Question

I'm reading Weibel's book on homological algebra right now and he's proving that for two $R$-modules $A$ and $B$, the equivalence classes of extensions of $A$ by $B$ (i.e. equivalence classes of short exact sequences $0\to B\to X\to A\to 0$). Fix a projective module $P$, a surjection $P\to A$ and let $M$ be such that $0\to M\to P\to A\to 0$ is exact (i.e. $M$ is the kernal of $P\to A$).

I want to show that the map $\Theta$ which sends an extension to the element $\partial(1_B)$ in $\mathrm{Ext}^1(A,B)$ (where $\partial$ is the connecting homomorphism the long exact sequence for $\mathrm{Ext}^*(-,B)$) is injective. He does this by constructing a left inverse $\Psi$ which I will construct now.

We have an exact sequence $$\mathrm{Hom}(P,B)\to \mathrm{Hom}(M,B)\rightarrow \mathrm{Ext}^1(A,B)\to 0$$ which comes from applying $\mathrm{Ext}^*(-,B)$ to the exact sequence $0\to M\to P\to A\to 0$ and using the fact that $P$ is projective. For $x\in \mathrm{Ext}^1(A,B)$ we lift to $\beta:M\to B$ and then let $X$ be the pushout of $B\leftarrow M\to P$. The map $X\to A$ is induces by $P\to A$ and the zero map $B\to A$. One can check that this actually yields an extension and only depends on the lift of $x$ up to isomorphism.

Now we want to prove that $\Psi$ is indeed a left inverse for $\Theta$. We take an extension and find its class in $\mathrm{Ext}^1(A,B)$ and lift it to $\gamma:M\to B$. Now I want to find a $\tau:P\to X$ which will make the right square below a pushout.

\begin{array}{ccccccccc} 0 & \xrightarrow{} & M & \xrightarrow{j} & P & \xrightarrow{} & A & \xrightarrow{} & 0\\ & & \downarrow{\gamma} & & \downarrow{\tau} & & \parallel & & \\ 0 & \xrightarrow{} & B & \xrightarrow{i} & X & \xrightarrow{} & A & \xrightarrow{} & 0 \end{array} Using projectivity of $P$ I can lift $X\to A$ to a $\tau:P\to X$. Weibel leaves it as an exercise to show that now $i$ and $\tau$ form the pushout of $\gamma$ and $j$.

Why is this true? This is the end of the proof which has been fine so far but I cannot for the life of me figure out why this square is a pushout. I don't even know why it commutes. I have no way of relating $\gamma$ and $\tau$ and so I'm at a loss. Any help, even just a hint for how to proceed, would be appreciated.

Answer 1

To show the square is a pushout, consider the map $\phi:P\oplus B\to X$ defined by $\tau$ and $P$ and $i$ on $B$. For the square to be a pushout, we need to prove that $\phi$ is surjective with kernel $\{(j(m),-\gamma(m):m\in M\}$.

This is diagram chasing. Write $\pi:X\to A$ and $\sigma:P\to A$ for the surjections in the diagram. For $x\in X$, $\pi(x)=\sigma(p)$ where $p\in P$. Then $\pi(x-\tau(p))=0$ and so $x-\tau(p)=i(b)$ where $b\in B$. Then $x=\phi(p,b)$: $\phi$ is surjective.

Let $(p,b)\in\ker\phi$. Then $\tau(p)=-i(b)$ and so $\sigma(p)=-\pi(i(b))=0$. Then $p=j(m)$ where $m\in M$. Also $i(\gamma(m))=\tau(j(m))=-i(b)$ and as $i$ is injective, $\gamma(m)=-b$. So $(p,b)=(j(m),-\gamma(m))$: we have the right kernel.

Answer 2

Suppose $Y$ is the pushout of $j : M \to P$ and $\gamma : M \to B$.

$$\require{AMScd} \begin{CD} 0 @>>>M @>>> P @>>> A @>>> 0\\ {} @VVV @VVV @VVV \\ 0 @>>>B @>>> Y @>>> A @>>> 0\\ \end{CD} $$

By universal property of pushout, there exists unique $w : Y \to X $.

And $w$ makes the diagram below commute. $$\require{AMScd} \begin{CD} 0 @>>>B @>>> Y @>>> A @>>> 0\\ {} @VVV @VVV @VVV \\ 0 @>>>B @>>> X @>>> A @>>> 0\\ \end{CD} $$

By 5-lemma, $w$ is an isomorphism.

Answer 3

We want to show that given an extension $E$ of $A$ by $B$, we have $\Theta(\Psi(E))=E$. Now as you said, $\Psi(E)=\delta(id_B)\in Ext_R^1(A,B)$ where $\delta$ is the connecting homomorphism obtained by applying $Ext_R^*(-,B)$ to our extension. Now $\Theta(\Psi(E))$ is the pushout extension $0\rightarrow B\rightarrow X\rightarrow A\rightarrow 0$, where $X$ is the pushout of

Where $\beta$ is any lift of $\delta(id_B)$ under the connecting homomorphism $\delta':Hom_R(K,B)\rightarrow Ext_R^1(A,B)$. The key here is to choose $\beta$ in a nice way.
Consider the diagram

where $s$ exists since $P$ is projective. Let $k\in K$. Then $\pi(\iota(k))=p(s(\iota(k)))=0$ and so $s(\iota(k))\in ker(p)=im(i)$. Hence there exists a unique $b\in B$ such that $s(\iota(k))=i(b)$. Thus we can define a map $$\beta:K\rightarrow B$$by letting $\beta(k)$ be the unique element $b\in B$ such that $s(\iota(k))=i(b)$. This completes the above diagram to give a commutative diagram

This is a morphism of short exact sequences and since $Ext_R^*(-,B)$ is a cohomological $\delta$-functor, we get a commutative square

from which we read off that $\delta'(\beta)=\delta(id_B)$ and hence the map we constructed, $\beta$, is indeed a valid lift!
Construct the pushout $X$ of this lift as mentioned at the begining. Then since we have forced the outer diagram below to commute, we get a (unique) map $\lambda$

which is given explicetely, by $\lambda\overline{(p,b)}=s(p)+i(b)$. Then by construction of the map $X\rightarrow A$ and all the commutation relations we have established so far, chasing along the maps we see that the following diagram commutes

and so from the $5$-lemma the result follows.