How to show $f^{-1}(D)/f^{-1}(C)\simeq D/C$ by the language of Abelian category?

Question

Let $\mathscr C$ be an Abelian category and $f:A\to B$ an epimorphism in $\mathscr C$.

Let $g:C\to D$ and $h:D\to B$ be monomorphisms in $\mathscr C$, thus $C,D$ are subobjects of $B$.

Let $f^{-1}(C)$ denote the inverse image of $C$ under $f$ and $f^{-1}(D)$ the inverse image of $D$ under $f$.

How to show $f^{-1}(D)/f^{-1}(C)\simeq D/C$ by the language of Abelian category?

Answer 1

A lemma I find very useful is:

Lemma: The square $$ \require{AMScd}\begin{CD} W @>u >> X \\ @VVV @VVV \\ Y @>v >> Z \end{CD} $$ is a pullback square if and only if both of the following conditions are true:

• The induced map $\ker(u) \to \ker(v)$ is an isomorphism
• The induced map $\operatorname{coker}(u) \to \mathrm{coker}(v)$ is monic

For intuition, note the second condition is equivalent to $\mathrm{coker}(u)$ being the image of the composite $X \to Z \to \mathrm{coker}(v)$.

The data you've given can be organized into the following diagram:

$$ \require{AMScd}\begin{CD} f^{-1}(C) @>>> f^{-1}(D) @>>> A \\ @VVV @VVV @VV f V \\ C @>>> D @>>> B \end{CD} $$

The square on the right and the outer square are pullback squares (by definition of $f^{-1}$); by the pasting lemma this means the square on the left is a pullback square as well.

By the above lemma, $f^{-1}(D) / f^{-1}(C) \to D/C$ is monic.

In an abelian category, pullbacks of epimorphisms are epimorphisms. This implies $f^{-1}(D) \to D$ is epic.

$D \to D/C$ is epic as well. Composites of epics are epic, and so the composite $f^{-1}(D) \to D \to D/C$ is epic as well.

But this composite is equal to $f^{-1}(D) \to f^{-1}(D)/f^{-1}(C) \to D/C$. In any category, if the composite $X \to Y \to Z$ is epic, then so is $Y \to Z$.

We've shown $f^{-1}(D)/f^{-1}(C) \to D/C$ is both monic and epic. In an abelian category, such morphisms must be isomorphisms.

Note that we did not need to use the fact $g$ and $h$ are monic. (although if we generalize, notation like $D/C$ should be understood as meaning the cokernel of the map $C \to D$)

Answer 2

By universal property of cokernels, there exists a morphism $h$ making the following diagram with exact rows commutative:

$$\require{AMScd}$$\begin{CD} O @>>> f^{-1}(C) @>>> f^{-1}(D) @>>> f^{-1}(D)/f^{-1}(C) @>>> O\\ {} @VVV @VVV @VV h V \\ O @>>> C @>>> D @>>> D/C @>>>O \end{CD}

Since $f$ is epic and since epimorphisms are pullback-stable, then $f^{-1}(D)\to D$ is an epimorphism. Consequently, $h$ is an epimorphisms as well. Since the left handed-square is a pullback, a diagram chasing show that $h$ is monomorphism, hence an isomorphism.