# Induced morphism between kernels is surjective

Let $$\mathcal{A}$$ be an abelian category. Given a commutative square $$\require{AMScd} \begin{CD} A@>f>>B\\ @VVV@VVV\\ A'@>>{f'}>B' \end{CD}$$ in $$\mathcal{A}$$, we get an induced morphism between kernels $$\operatorname{Ker}f\to\operatorname{Ker}f'$$, namely, the unique one that makes the diagram $$\require{AMScd} \begin{CD} \operatorname{Ker} f@>>>A\\ @VVV@VVV\\ \operatorname{Ker}f'@>>{}>A' \end{CD}$$ commutative. I was trying to prove the following

Lemma. Let $$\require{AMScd} \begin{CD} A\times_CB@>{p}>>A\\ @V{q}VV@VV{f}V\\ B@>>{g}>C \end{CD}$$ be a cartesian square in $$\mathcal{A}$$. If $$h:D\twoheadrightarrow A\times_CB$$ is a surjection, then the induced morphism between kernels from the commutative square $$\require{AMScd} \begin{CD} D@>>>A\\ @VVV@VV{f}V\\ B@>>{g}>C \end{CD}$$ is also surjective.

For me “surjective morphism” in an abelian category means an epimorphism or a morphism with zero cokernel (these two conditions are equivalent).

The proof is easily done in $$\mathcal{A}=R\operatorname{-Mod}$$ the category of left $$R$$-modules: if $$b\in B$$ is such that $$g(b)=0$$, then $$(0,b)\in A\times_CB$$, and there is $$d\in D$$ such that $$h(d)=(0,b)$$. Thus $$d\in\operatorname{Ker}(D\to A)$$. Also, in particular $$d$$ maps to $$b$$ in $$B$$. How one would do the proof in general? Is it okay to invoke Mitchell's embedding theorem or is it an overkill because there is an easier way? I don't exactly know how to translate the preceding element-wise diagram-chasing to a morphism-wise chasing using universal properties.

Edit: Okay this is what I've come up with so far. Using the universal property of the fibre product, there exists a morphism $$\operatorname{Ker}g\to A\times_C B$$ such that in the diagram

the composite $$\operatorname{Ker}g\to A\times_CB\xrightarrow{q} B$$ equals $$\operatorname{Ker}g\to B$$ and the composite $$\operatorname{Ker}g\to A\times_CB\xrightarrow{p}A$$ vanishes. Moreover, the morphism $$\operatorname{Ker}g\to A\times_CB$$ has to injective since after post-composition with $$q$$ we get an injective morphism $$\operatorname{Ker}g\to B$$. Also we have that the top left triangle of the diagram commutes, since equality of the two sides of the triangle can be verified after post-composition with $$p$$ and $$q$$.

But now I don't know how to continue. I haven't used the surjectivity of $$h$$ yet, and I don't know how to use it exactly.

For those who may ask why am I interested on this result: I'm trying to understand the proof of 05T7 of the Stacks Project. The proof is done by induction, and the verification of the induction hypothesis $$IH_{n-1}$$, which “the reader easily checks,” require to verify that an induced morphism between kernels is surjective. The above lemma abstracts the situation from the proof.

I know that there are posts here on MSE that address the same result I've linked from the Stacks Project, like this one. However, this last post uses other strategy, and I was interested on understanding the one from the SP.

Edit: the answer is essentially the same, but I simplified it a little, added details, and a new picture. I have also slightly changed my notations to better adapt them to yours.

Consider the following commutative diagram with exact rows, where the lower right square is your pullback and $$h$$ is your epimorphism:

In the notation of the above picture, you are asking to prove that the composition $$f''\circ p$$ is an epimorphism. Of course, it is enough to show that both $$f''$$ and $$p$$ are epimorphisms:

• $$f''$$ is an epimorphism: note that the maps $$k\colon \ker(g)\to B$$ and $$0\colon \ker(g)\to A$$ are such that $$g\circ k=0=f\circ 0$$ so that, by the universal property of the pullback, there is a unique morphism $$\bar k\colon \ker(g)\to A\times_CB$$ such that $$f'\circ \bar k=k$$ and $$g'\circ \bar k=0$$. The latter condition tells us that, by the universal property of the kernel, there is a unique morphism $$\tilde k\colon \ker(g)\to \ker(g')$$ such that $$k'\circ\tilde k=\bar k$$. Then, $$k\circ f''\circ \tilde k=f'\circ k'\circ \tilde k=f'\circ \bar k=k=k\circ \mathrm{id}_{\ker(g)},$$ which shows that $$f''\circ \tilde k=\mathrm{id}_{\ker(g)}$$ since $$k$$ is a monomorphism. In particular, $$f''$$ is an epimorphism. (In fact, $$f''$$ is an isomorphism, see The Stacks project, Tag: 08N3, with inverse $$\tilde k$$, but this is not needed here).

• $$p$$ is an epimorphism: just note that $$\ker(g'')=h^{-1}(\ker(g'))$$, that is, the upper left square is also a pullback, and epimorphisms are pullback stable (see, e.g., The Stacks project, Tag: 08N4).

• In case you haven't seen it: shortly before you posted your answer, I added an edit where I did the same thing as you did on your second paragraph Jun 23, 2022 at 16:09
• Yes, I saw it after posting. The moral is the following: your argument (=the argument in the second paragraph of the answer) proves the lemma for the epimorphism $id_{A\times_CB}\colon A\times_CB\to A\times_CB$, that is, the induced map $\ker(g')\to \ker(g)$ is epic. Now, as any other epimorphism $p\colon P\to A\times_CB$ factors through $id_{A\times_CB}$, you have that the kernel of $g'\circ p$ always maps onto $\ker(g')$, which maps onto $\ker(g)$ by the first part. Jun 23, 2022 at 16:14
• If you need more details, just let me know and I will edit the answer with the required explanations! Jun 23, 2022 at 16:17
• Thanks! You're using 08N4, right? Jun 23, 2022 at 16:27
• Sorry, now I see what you meant by "using 08N4"! Yes, that gives you that $p_{\restriction p^{-1}(0\times_C\ker(g))}$ is epic. Jun 23, 2022 at 16:52