Question in the proof of theorem 2.2.2 in Grothendieck article's

Question

I'm reading the article "Sur quelques points d'algèbre homologique" of Grothendieck (here a pdf http://matematicas.unex.es/~navarro/res/tohoku.pdf) and I'm stuck at the theorem 2.2.2. In the proof, he defines $S^1F(A) = F(A')/Im F(M)$ by using an exact sequence and he says that it doesn't depend on the choice of the exact sequence since another choice of exact sequence will lead to another $S^1F(A)$ which will be isomorphic. I don't see how to prove this, I took a look at the reference he called but It is not clear for me. Does someone see how to do this please ?

Answer 1

$\require{AMScd}$Let $\mathcal C$ be an abelian category such that every object $A$ in $\mathcal C$ admits an injective effacement, that is, there exists a monomorphism $A \to M$ (called an injective effacement) such that for all monomorphisms $B\to C$ and all morphisms $B\to A$ there exists a morphism $C\to M$ making the diagram $$ \begin{CD} B @>>> C \\ @VVV @VV{\exists}V \\ A @>>> M \end{CD} $$ commutative.

Let now $F\colon \mathcal C \to \mathcal D$ be a covariant additive functor, where $\mathcal D$ is another abelian category. The first satellite $S^1F$ is defined as follows: let $A \in \mathcal C$ and consider the short exact sequence $0\to A\xrightarrow{i} M\xrightarrow{\pi} A'\to 0$, where $i$ is an injective effacement. Then $S^1F(A)$ is defined by the exact sequence $$ F(M) \longrightarrow F(A') \longrightarrow S^1F(A) \longrightarrow 0. $$ We need to check that this definition is independent of the choice of injective effacement and that $S^1F$ is a functor. We do both at once.

Consider a commutative diagram $$ \begin{CD} 0 @>>> A @>i>> M @>{\pi}>> A' @>>> 0 \\ & @V{\alpha}VV @V{\varphi}VV @VV{\beta}V \\ 0 @>>> B @>>j> N @>>{\rho}> B' @>>> 0, \end{CD} $$ where $j\colon B\to N$ is an injective effacement, which induces a (non-unique) map $\varphi$ making the left square commutative; the map $\beta$ is uniquely determined by the universal property of the cokernel $\pi\colon M\to A'$. Applying $F$, we obtain a commutative diagram $$ \begin{CD} F(M) @>{F(\pi)}>> F(A') @>>> S^1F(A) @>>> 0 \\ @V{F(\varphi)}VV @VV{F(\beta)}V @VV{S^1F(\alpha)}V \\ F(N) @>>{F(\rho)}> F(B') @>>{\tau}> S^1F(B) @>>> 0. \end{CD} $$ where the map $S^1F(\alpha) \colon S^1F(A) \to S^1F(B)$ is the unique map determined by the universal property of the cokernel $F(A') \to S^1F(A)$.

Claim: If $\alpha = 0$, then for any choice of $\varphi$, we have $S^1F(\alpha) = 0$.
Proof: Suppose $\alpha = 0$. Then $\varphi\circ i = 0$, hence by the universal property of the cokernel $\pi\colon M\to A'$, there exists $s\colon A'\to N$ such that $\varphi = s\circ\pi$. Note that $\rho\circ s\circ\pi = \rho\circ \varphi = \beta\circ \pi$, and since $\pi$ is an epimorphism, this implies $\beta = \rho\circ s$. But now $$ \tau \circ F(\beta) = \tau\circ F(\rho)\circ F(s) = 0, $$ from which we deduce $S^1F(\alpha) = 0$. The claim is proved. $\qquad\Box$

By additivity, the claim shows that for any $\alpha\colon A\to B$, the map $S^1F(\alpha)$ is independent of the choice of $\varphi$. In other words, $S^1F(\alpha)$ is canonical.

It is now a trivial exercise to check $S^1F(\alpha_1\circ\alpha_2) = S^1F(\alpha_1)\circ S^1F(\alpha_2)$ and $S^1F(\operatorname{id}_A) = \operatorname{id}_{S^1F(A)}$.