question about commutative diagram in category theory

Question

I am reading the article

• Maurice Auslander, Ragnar-Olaf Buchweitz, The homological theory of maximal Cohen-Macaulay approximations, Colloque en l'honneur de Pierre Samuel (Orsay 21-22 mai 1987), Mémoires de la Société Mathématique de France, Série 2, no. 38 (1989), pp. 5-37. doi:https://doi.org/10.24033/msmf.339

In this article, the proof Lemma 3.1 constructs a commutative diagram as follows. We work in an Abelian category $\mathbf{C}$ with a full, additively closed, exact subcategory $\mathbf{X}$ (and a few more hypotheses that don’t seem to be relevant). Given exact sequences $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ $$0 \rightarrow Y_A \rightarrow X_A \rightarrow A \rightarrow 0$$ where the latter is an $\mathbf{X}$-approximation of $A$, the proof says that since $$Ext^1(C,X_A) \rightarrow Ext^1(C,A)$$ is an isomorphism, there exists an exact commutative diagram like:

My question is: How to calculate Z? Is it a pull back or push out?

(P.S: The above is in the context of an abelian category, in which the existence of projective and/or injective objects is not assumed.)

Answer 1

Choose a projective module $P$ and a surjective homomorphism $\epsilon\colon P\to C$, with kernel $K$ say. Then $\text{Ext}^1(C,M)=\text{Hom}(K,M)/i^*\text{Hom}(P,M)$ for all $M$. As $P$ is projective and the map $B\to C$ is surjective, we can choose $P\to B$ giving a commutative triangle. This will restrict to give a homomorphism $K\to A$. This represents a class in $\text{Ext}^1(C,A)$, and by assumption that must come from a class in $\text{Ext}^1(C,X_A)$, and thus from a morphism $K\to X_A$. You can then take $Z$ to be the pushout of $X_A\xleftarrow{}K\xrightarrow{}P$. As $K\to P$ is injective with cokernel $C$, we see that $X_A\to Z$ is also injective with cokernel $C$. There are some further steps needed to complete the commutative diagram, but I will leave those to you rather than attempting to draw the relevant diagrams here.

Answer 2

We may work in the derived category $D^b(\mathbf{C})$. The given map $\pi\colon X_A\to A$ induces a bijection $$\mathrm{Hom}_{D^b(\mathbf{C})}(C,X_A[1])\cong\mathrm{Ext}^1(C,X_A) \cong\mathrm{Ext}^1(C,A)\mathrm{Hom}_{D^b(\mathbf{C})}(C,A[1])\, .$$ Interpreting the short exact sequence $0\to A\to B\to C\to 0$ as a map $f\colon C\to A[1]$ in ${D^b(\mathbf{C})}$, there is a unique map $g\colon C\to X_A[1]$ such that $(\pi[1])\circ g=f$. We can extend the map $g$ into a distinguished triangle $$X_A\to Z\to C\to X_A[1]\, .$$ The identity of $C$ and the map $\pi$ may be completed to a morphism of triangles from the latter to the distinguished triangle $$A\to B\to C\to A[1]\, .$$ By the octaedral axiom, and interpreting $\mathbf{C}$ as the full subcategory of ${D^b(\mathbf{C})}$ spanned by objects whose chomology is concentrated in degree zero, we get the commutative diagram above, with exact rows and columns.

For those who prefer derived $\infty$-categories, one may replace the octahedral axiom by basic facts on pull-back squares in $\infty$-categories, which makes the construction even more straightforward.

Whatever recipy we find to construct $Z$, it will involve non-canonical choices in $\mathbf{C}$. In fact, it is more useful to understand how to recover $B$ from $Z$: the identification $\mathrm{Ext}^1(C,X_A)\cong\mathrm{Ext}^1(C,A)$ is obtained by sending any short exact sequence $$0\to X_A\to Z\to C\to 0$$ to the short exact sequence $$0\to A\to B'\to C\to 0$$ where $B'$ is the push-out of $Z$ along the map $X_A\to A$, and that explains everything we need to know in the paper of Auslander and Buchweitz: there exists $Z$ such that $B\cong B'$ holds.