# Second Isomorphism Theorem in arbitrary abelian category

I would like to prove the Second Isomorphism Theorem in an arbitrary abelian category without appealing to Mitchell's Embedding Theorem.

Given two subobjects $M', M''$ of an object $M$ in an abelian category $\mathcal A$, we define the intersection $M'\cap M'' := \ker(M\rightarrow M/M' \oplus M/M'')$ and the sum $M'+M'' := \text{im}(M'\oplus M''\rightarrow M)$, where these morphisms are all canonical.

I want to show that $(M'+M'')/M' \cong M''/M'\cap M''$.

We get a map $M'' \rightarrow (M'+M'')/M'$ by injecting $M''$ into $M'\oplus M''$ and using the universal property of the image to get a map to $M'+M''$ and then composing with the projection.

I don't know where to go from here though. I know that for any map $f$ we have $\text{coker}\ker f = \ker\text{coker} f$, and I realise I probably need to use this since the first isomorphism theorem is used to prove the second in the case of modules, but I'm stuck and all the arrows are confusing me.

For an alternative proof without using the snake lemma : consider the diagram $$\require{AMScd} \begin{CD} M' \cap M'' @>>> M'' @>>> \frac{M''}{M' \cap M''} @>>> 0 \\ @VVV @VVV @VVV \\ M' @>>> M' + M'' @>>> \frac{M'+M''}{M'}@>>> 0.\end{CD}$$ As Hurkyl explained, the left-hand side square is a pushout. Now by definition of cokernel so are the squares \begin{CD}M'@>>> M'+M'' \\ @VVV @VVV \\ 0@>>> \frac{M'+M''}{M'} \end{CD} and thus the outer square in the rectangle \begin{CD} M' \cap M'' @>>> M' @>>> 0 \\ @VVV @VVV @VVV \\ M'' @>>> M' + M'' @>>> \frac{M'+M''}{M'}\end{CD} is a pushout. But this is the same as the rectangle \begin{CD} M' \cap M'' @>>> 0 @>>> 0 \\ @VVV @VVV @VVV \\ M'' @>>> \frac{M''}{M'\cap M''} @>>> \frac{M'+M''}{M'},\end{CD} and the left-hand square is again a pushout. Thus the right-hand side is a pushout, and thus the arrow $\frac{M''}{M'\cap M''}\to\frac{M'+M''}{M'}$ is an isomorphism.

Two alternate formulas simplify this problem.

The intersection of subobjects is more commonly defined to be the pullback

$$\require{AMScd} \begin{CD} M' \cap M'' @>>> M' \\ @VVV @VVV \\ M'' @>>> M \end{CD}$$

which, in an abelian category, is

$$M' \cap M'' \cong \ker( M' \oplus M'' \to M)$$

In nice categories, the join of subobjects can be given by the pushout

$$\require{AMScd} \begin{CD} M' \cap M'' @>>> M' \\ @VVV @VVV \\ M'' @>>> M' + M'' \end{CD}$$

which, in an abelian category, is

$$M' + M'' \cong \operatorname{coker}(M' \cap M'' \to M' \oplus M'')$$

which can be seen to be equivalent to the formula you give, by the fact images are coimages.

Using these alternative characterizations may simplify the problem for you.

I would continue as follows. The map in the other direction is now easy to give: constructing the commutative square on the left gives a transformation of cokernel diagrams

$$\begin{CD} & & M' \oplus (M' \cap M'') @>\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}>> M' \oplus M'' @>>> (M' + M'') / M' @>>> 0 \\ & @VV\begin{pmatrix}0 & 1\end{pmatrix}V @VV\begin{pmatrix}0 & 1\end{pmatrix}V @VVV \\ 0 @>>> M' \cap M'' @>>> M'' @>>> M'' / (M' \cap M'') @>>> 0 \end{CD}$$

(I've used matrix notation to indicate the map you construct from and to direct sums via the individual components. The notation convention is to treat the direct sum like column vectors)

The snake lemma quickly proves the rightmost vertical map is an isomorphism. (and also that the top row is short exact as well)

I suppose I haven't shown that the cokernel of the top row is the claimed group; I assume that is a straightforward exercise.