# Cocartesian squares in the category of abelian groups.

Recently, I've been doing a recap of (basic) category theory and found an old exercise I seem to be unable to solve. The question is as follows.

Let $A, B$ be abelian groups, $A'<A$ and $B'<B$ subroups and $\phi:A\to B$ such that $\phi(A')\subseteq B'$. Let $\phi':A'\to B'$ and $\phi'':A/A'\to B/B'$ be the maps induced by $\phi$. Prove that the diagram

\begin{array} AA' & \to & A \\ \downarrow{\phi'} & & \downarrow{\phi} \\ B' & \to & B \end{array}

is a cocartisian square if and only if $\phi''$ is an isomorphism.

The exercise preceding this one looks like a dual statement which I did manage to prove (a similar square being cartesian iff $\phi'$ is an iso), yet just dualising the proof doesn't seem to work. That proof, however, suggests using the universal property of cocartesian squares for one of the implications ($\Rightarrow$), while using the explicit construction of pushouts in the category of abelian for the other one. The pushout of $A_1\overset{f_1}{\leftarrow} A_3 \overset{f_2}{\rightarrow} A_2$ in this category is given by $A_1\oplus A_2/\{(f_1(x),-f_2(x))\mid x\in A_3\}$.

Can anyone provide me with a proof, or give a sketch of proof that might point out which step I fail to see just yet?

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Wow, this is basic? :) – rschwieb May 21 '13 at 21:08
Well, it occured in the third or fourth week in a second-year course on category theory. Even though the course was far too difficult at the time, I assumed this would still make it basic category theory. I'm willing to take out the word 'basic' from the description, if you think that suits it better. – HSN May 21 '13 at 21:11
@rschwieb, it's another (weird) name for a pushout, which is the categorical dual of a pullback, which is also sometimes known as "fibered product". – Andy May 22 '13 at 8:13
@Andy "Cocartesian" does not name the pushout (i.e. the vertex of the colimit), but the diagram involving the pushout. So it is a name for the universal cocone with vertex the pushout. (This designation comes from french, probably from EGA/SGA.) – Pece May 22 '13 at 12:24
@Andy Oh yes you're absolutely right. I assumed too fast that you were talking about the vertex. Sorry for the misunderstanding. – Pece May 22 '13 at 12:43

## 1 Answer

Take $A' \to A \to A/A'$, the first map is the inclusion and the second is the quotient projection. Take then $B' \to B \to B/B'$, the maps as above. These are both exact sequences, which means $\operatorname{im}(\text{inclusion}) = \ker(\text{projection})$. You should notice that $(\phi', \phi, \phi'')$ is a morphism of exact sequences, which means that this triple makes the obvious diagram commute.

Now, you may check that the sequence $B' \to P \to \operatorname{coker}(i)$ is exact, if $P$ is the pushout of the inclusion $A' \subseteq A$ along $\phi'$ and $i$ is the standard inclusion of $B'$ into the direct sum you mention. If you put the two diagrams together (the one that describes the map $(\phi', \phi, \phi'')$ and the one with $(\phi', j, k)$, where $j\colon A \to P$ is the standard inclusion and $k\colon A/A' \to \operatorname{coker}(i)$), you notice that there is also a map form $B$ to $P$ because of the universal property of $P$. As you should know the universal map is an iso, so $B \cong P$. Then you notice that $\operatorname{coker}(i)$ is iso to $B/B'$, because the inclusions $B' \subseteq B$ and $B' \subseteq P$ differ by an iso. This means that your square is in fact a pushout.

This is a succession of iff-s, so your claim is proved. This should be so much easier if you write down the diagrams I mention, which I cannot do here, since it looks like mathjax does not support xypic, and I'm not familiar with any other way of writing commutative diagrams.

Clarification: I understand this is an exercise in category theory and I didn't use any strictly categorical methods, but you should notice that the category of abelian groups is an abelian category and this is actually done in utter generality in any abelian category if, instead of quotients you take cokernels.

EDIT: Here's the diagram I was talking about.

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You can try presheaf for drawing diagrams (it's probably best to right click the rendered diagram, copy the link and upload the image into your answer). – Martin May 21 '13 at 23:10
@Martin, very nice thanks! I'll edit the question tomorrow morning, it's kind of late right now. – Andy May 21 '13 at 23:11
@Andy: I unfortunately hadn't been able to look at this before. Up to the point where you say 'so $B\cong P'$ I can follow you, but I fail to see why this iso holds. It is clear where the unique map $B\to P$ comes from. I'd, however, only expect it to be an iso if we already knew $B$ to be the pushout, which is what we're trying to prove. What is it that I'm still missing? – HSN May 28 '13 at 10:24
@Martin: thanks a lot! I had problems writing diagrams as well. – HSN May 28 '13 at 10:25
@HSN: the universal morphism is always an iso if all the right squares commute, because it means that the other object is also a (co)limit. Try to write it out in general: take a limiting (co)cone and another limiting (co)cone (this means that everything that needs to commute, does) you have two unique morphisms one from and one to; it should be easy to convice yourself that these are one the inverse of the other. This is the case in this situation: all the "right" squares commute. – Andy May 28 '13 at 10:46