# Why the rightmost square is a pullback?

I am currently stuck in the problem below, which is somehow a converse to a previous question I asked.

Suppose that the following commutative diagram in an abelian category $$\require{AMScd} \begin{CD} 0 @>>> X'@>>>X@>{g}>>X''@>>>0\\ @.@V{f'}VV @VV{f}V @VV{f''}V\\ 0@>>> Y'@>>>Y@>{h}>>Y''@>>>0 \end{CD}$$ has exact rows and $f'$ is an isomorphism. Then the rightmost square $\require{AMScd} \begin{CD} X@>>>X''\\ @VVV @VVV\\ Y@>>>Y'' \end{CD}$ is a pullback.

I think the snake lemma could help in this case, which yields that $f$ and $f''$ are monic and thus it also suffices to show the square is a pushout (it seems easier to check...) What I actually have so far is no more than this and I would like to ask for some hints about what to do next.

Any help is appreciated.

• Can you come up with a diagram-chasing argument which would work in the case of the category of abelian groups, for example? If so, then the previous answer of mine here could give a method for transforming this into an argument working in arbitrary abelian categories: math.stackexchange.com/questions/2354953/… Mar 26 '18 at 19:02
• Also, off the top of my head, I would think the snake lemma would just give an isomorphism between the kernels of $f$ and $f''$, and similarly an isomorphism between their cokernels. Mar 26 '18 at 19:06
• @DanielSchepler Yeah, you are right. Snake lemma only gives an isomorphism between their kernels... For the moment I have not worked out a diagram-chasing argument yet, but I am very curious that, in the link you gave me, the accepted answer seems to have provided an argument to prove my problem. To be specific, that answer asserts that "because the kernels are isomorphic, the square $*$ is a pullback". I wonder this is enough to prove my question with some minor adjustment? Mar 26 '18 at 19:29
• That other answer seems to be making use of the fact you're trying to prove here. Mar 26 '18 at 19:54
• @DanielSchepler At that moment I was thinking whether there were insights that could lead to the proof with such a single argument in that answer, lol. Mar 27 '18 at 1:07

To prove that the rightmost square is a pullback, it is enough to prove that the canonical arrow $(f,g):X\to Y\times_{Y''}X''$ (which exists because the square commutes) is an isomorphism. To this end, look at the decomposition of your square into two commutative squares, one induced by the equality $\psi_2\circ (f,g)=g$ and the other one given by the actual pullback $Y\times_{Y''}X''\to X''$. Then taking kernels yields a commutative diagram $$\require{AMScd} \begin{CD}0 @>>> X' @>{\ker(g)}>> X @>{g}>> X'' @>>> 0 \\ & @V{u}VV @V{(f,g)}VV @VV{1_{X''}}V \\ 0 @>>> K @>{\ker(\psi_2)}>> Y\times_{Y''}X'' @>{\psi_2}>> X'' @>>> 0 \\& @V{v}VV @V{\psi_1}VV @VV{f''}V \\ 0 @>>> Y' @>>{\ker(h)}> Y @>>{h}> Y'' @>>> 0 \end{CD}$$ whose rows are exact ($\psi_2$ is an epi because $g$ is). Here $v$ is the factorisation of $\psi_1\circ \ker(\psi_2)$ through $\ker(h)$, which must exist since $$h\circ \psi_1\circ \ker(\psi_2)=f''\circ \psi_2 \circ \ker(\psi_2)=0;$$ and $u$ is the factorization of $(f,g)\circ \ker(g)$ through $\ker(\psi_2)$, which must exist since $$\psi_2\circ (f,g)\circ \ker(g)=g\circ \ker(g)=0.$$ Notice that $$\ker(h)\circ v\circ u= \psi_1\circ \ker(\psi_2)\circ u=\psi_1\circ (f,g)\circ \ker(g)=f\circ \ker(g)=\ker(h)\circ f',$$ and thus $v\circ u=f'$ since $\ker(h)$ is a monomorphism. Now the result in your previous question shows that $v$ must be an iso, so that if $f'$ is an iso, so is $u$. Then applying the Short Five Lemma to the upper part of the diagram shows that $(f,g)$ is an isomorphism.
• May I ask one more question: is $K$ the kernel of $\psi_2$ and how such $u$, $v$ are found, by universal properties? According to your MO answer, it seems that $K$ here can be replaced by $Y'$ with $u=f'$ and $v=\mathrm{id}_{Y'}$ and the proof also works. I mean if we already have the argument that $f'=vu$ then the commutative diagram you gave in the MO answer follows but how $u$ and $v$ can be constructed in advance in the proof above? Mar 27 '18 at 1:30
• @josephz Yes, $K$ is the kernel of $\psi_2$; I've added some explanations on how $u$ and $v$ are defined. The argument in the MO answer is slightly different : there I replaced the kernel of $\psi_2$ by the map $(\ker(h),0)$ given by the universal property of the pullback and the fact that (in your notation) $h\circ \ker(h)=0=\psi_2\circ 0$. Then you get the two commutative squares on the left with $v=1_{Y'}$ and $u=f'$, but you need to prove that $(\ker(h),0):Y'\to Y\times_{Y''}X''$ is a kernel of $\psi_2$. Mar 27 '18 at 10:18
• Thanks for these explanations! I feel much clearer now. However, I am also a little confused that, I guess the notation $\require{AMScd} \begin{CD} Y\times_{Y''}X''@>{\psi_2}>>\ \\ @V{\psi_1}VV @.\\ \ @. \ \end{CD}$ is the pullback constructed in allusion to $\require{AMScd} \begin{CD} \ @. X''\\ @. @VVV\\ Y@>>> Y''\end{CD}$ through the product $Y\times X''$ (which is possible because an abelian category admits all finite products and kernels)? And in light of this how can $\psi_2$ be defined to ensure $\psi_2\circ (f,g)=g$ (and analogously, for $\psi_1$)? Mar 27 '18 at 12:26
• @josephz The pullback $Y\times_{Y''}X''$ is a subobject of the product $Y\times X''$. You can obtain it as the kernel of the map $\operatorname{coker}((1_{Y''},1_{Y''}))\circ (h\times f''): Y\times X''\to Coker((1_{Y''},1_{Y''}))$ (note that if you know that your category is additive, then this map is equivalently given by $h\pi_1-f''\pi_2$). Mar 27 '18 at 12:50
• Um...yeah, this is how the pullback is constructed, but how can we say that $$\psi_2\circ(f,g)=g,\ \psi_1\circ(f,g)=f$$ is guaranteed? Mar 27 '18 at 14:22