Pushforward commutes with pullback in short exact sequences

Question

In 010I we find the definition of the pullback (resp., the pushforward) of a short exact sequence $0\to A\to E\to B\to 0$ in some abelian category along a morphism $B'\to B$ (resp., along a morphism $A\to A'$), giving rise to a s.e.s. $0\to A\to E\times_B B'\to B'\to 0$ (resp., to a s.e.s. $0\to A'\to A'\amalg_A E\to B\to 0$). Then, it is claimed that $$ \tag{1}\label{iso} A' \amalg_A (E \times_B B') \cong (A' \amalg_A E)\times_B B', $$ i.e., that it doesn't matter if we take the pullback first and after the pushforward or viceversa; we will end up with the same s.e.s.

My question is: how do we prove \eqref{iso}? I've tried to build some natural morphism from the LHS to the RHS by appealing to universal properties of pushouts, but I have failed to achieved this.

The Stacks Project say:

Recall that $A' \amalg_A E$ is a quotient of $A' \oplus E$. Thus the right hand side is a quotient of $A' \oplus E \times_B B'$, and it is straightforward to see that the kernel is exactly what you need in order to get the left hand side.

But I don't understand this: how $(A' \amalg_A E)\times_B B'$ is a quotient of $A' \oplus E \times_B B'$? (Do they actually mean $(A' \oplus E )\times_B B'$?)

Answer 1

I just found the result in Mac Lane's Homology, III, Lemma 1.6 (basically the same proof is done in B. Mitchell, Theory of Categories, VII, Lemma 1.3.iii). For reference, I will sum up his proof here (he does it for $R$-modules but we shall do it over an abelian category). The following numbering of the lemmas corresponds to that in Mac Lane's book. Let's introduce Mac Lane's notations: Fix an abelian category inside which all the upcoming sequences will live. Given a short exact sequence $$ \tag{2}\label{ses} 0\to A\to B\to C\to 0, $$ which we denote as $E$, and a morphism $\gamma:C'\to C$, a pullback of $E$ along $\gamma$ is any morphism of s.e.s. $(1,\beta,\gamma):E\gamma\to E$.

Lemma 1.3. A pullback of $E$ along $\gamma$ exists, is unique up to unique isomorphism over $E$ and satisfies the following universal property: given a s.e.s. $(\alpha_1,\beta_1,\gamma_1):E_1\to E$ over $E$ with $\gamma_1=\gamma$, there is a unique morphism $\beta':B_1\to B\gamma$ (where $B_1$ and $B\gamma$ denote, respectively, the middle terms of $E_1$ and $E\gamma$) such that $(\alpha_1,\beta',1):E_1\to E\gamma$ is a morphism and such that the composite $$ E_1\xrightarrow{(\alpha_1,\beta',1)}E\gamma\xrightarrow{(1,\beta,\gamma)}E $$ equals $E_1\to E$.

Proof. Define $B'=B\times_CC'$, so that we have a morphism $$ \tag{3} \label{pullback} \require{AMScd} \begin{CD} 0@>>>A@>>>B'@>>>C'@>>>0\\ @.@|@VVV@VV{\gamma}V@.\\ 0@>>>A@>>>B@>>>C@>>>0 \end{CD} $$ of s.e.s. (the top row is exact by 08N3 and 08N4). This shows existence. Note that the uniqueness follows from the universal property, for if $\alpha_1=1$, then by the five lemma the morphism $(\alpha_1,\beta',1):E_1\to E\gamma$ is an isomorphism. From the presentation \eqref{pullback} of $E\gamma\to E$ it is an easy diagram chase to show the universal property. $\square$

In other words, $E\gamma\to E$ is final in the non-full subcategory of the s.e.s. over $E$ whose objects are $(\alpha_1,\beta_1,\gamma):E_1\to E$ and whose morphisms over $E$ are of the form $(\alpha',\beta',1):E_1\to E_2$ (the object $E\gamma\to E$ is weakly final—but in general not final—in the full category spawned by the s.e.s. over $E$ of the form $(\alpha_1,\beta_1,\gamma):E_1\to E$, since to build a factorization $(\alpha',\beta',\gamma'):E_1\to E\gamma$ of $E_1\to E$ we need to impose $\alpha'=\alpha_1$, but then we may pick any $\gamma':C'\to C'$ such that $\gamma=\gamma\gamma'$).

Consider again $E$ as in \eqref{ses} and a morphism $\alpha:A\to A'$. A pushout of $E$ along $\alpha$ is any morphism $(\alpha,\beta,1):E\to\alpha E$ of s.e.s.

The following is the dual of Lemma 1.3:

Lemma 1.5. A pushout of $E$ along $\alpha$ exists, is unique up to unique isomorphism under $E$ and satisfies the following universal property: given a s.e.s. $(\alpha_1,\beta_1,\gamma_1):E\to E_1$ under $E$ with $\alpha_1=\alpha$, there is a unique morphism $\beta':\alpha B\to B_1$ (where $\alpha B$ and $B_1$ denote, respectively, the middle terms of $\alpha E$ and $E_1$) such that $(1,\beta',\gamma_1):\alpha E\to E_1$ is a morphism and such that the composite $$ E\xrightarrow{(\alpha,\beta,1)}\alpha E\xrightarrow{(1,\beta',\gamma_1)}E_1 $$ equals $E\to E_1$.

Thus, $E\to E\alpha$ is initial in the non-full subcategory of the s.e.s. under $E$ whose objects are $(\alpha,\beta_1,\gamma_1):E\to E_1$ and whose morphisms under $E$ are of the form $(1,\beta',\gamma'):E_1\to E_2$.

Lemma 1.6. For $\alpha$ and $\gamma$ as in Lemmas 1.2 and 1.4 there is a canonical isomorphism $\alpha(E\gamma)\cong(\alpha E)\gamma$.

That is, we have \eqref{iso}.

Proof. The composite morphism $$ E\gamma\xrightarrow{(1,\beta_1,\gamma)}E\xrightarrow{(\alpha,\beta_2,1)}\alpha E $$ is $(\alpha,\beta_2\beta_1,\gamma):E\gamma\to\alpha E$. Using Lemmas 1.3 and 1.4, we get a solid diagram:

where both outer composites equal $(\alpha,\beta_2\beta_1,\gamma)$. We are done, since $(1,\beta',\gamma): \alpha(E\gamma)\to \alpha E$ is a pullback of $\alpha E$ along $\gamma$ (or $(\alpha,b',1):E\gamma\to (\alpha E)\gamma$ is a pushout of $E\gamma$ along $\alpha$). Thus, there exists a unique dashed isomorphism making the whole diagram commute. $\square$

This justifies the usage of the notation $\alpha E\gamma$ for the isomorphism class of $\alpha(E\gamma)\cong(\alpha E)\gamma$. In particular, the Baer sum of $E_1,E_2\in\operatorname{Ext}(C,A)$ can be defined as $E_1+E_2=\nabla(E_1\oplus E_2)\Delta$, where $\Delta:C\to C\oplus C$ is the diagonal morphism and $\nabla:A\oplus A\to A$ is the codiagonal morphism.