Why $A \hookrightarrow B \hookrightarrow C \simeq A$ implies $A \simeq B$? Why is it true that (at least in an abelian category) if we have arrows $$A \hookrightarrow B \hookrightarrow C \simeq A$$ then $A \simeq B$? This seems like a categorical version of Cantor-Bernstein but I am not able to justify this formally.
(This question is motivated by the fact that this is used on page 178 of Kashiwara-Schapira, Categories and sheaves, where a null sequence of two arrows $X' \xrightarrow{f} X \xrightarrow{g} X''$ is considered; in this case $A=\mathrm{im} f$, $B= \ker u$, $C=\ker(X \to \mathrm{coker} f)$ with $u=\ker g \to X \to \mathrm{coker} f$.)
 A: It is not true in a general abelian category that if $A$ is a subobject of $B$ and $B$ is a subobject of $A$ then $A\simeq B$.
The reason it works in the context considered by Kashiwara and Schapira is that $\mathrm{im} f$, $\ker u$ and $\ker(X \to \mathrm{coker} f)$, and the morphisms between them, are not simply considered as objects but more specifically as subobjects, and inclusions between subobjects, of $X$. In particular, it implies that each morphism in the composition
$$\mathrm{im} f \hookrightarrow \ker u \hookrightarrow \ker(X \to \mathrm{coker} f) \simeq \mathrm{im} f$$
commutes with each subobject's monomorphism into $X$. This means that the composition above must in fact be the identity on $A$; in particular, $\ker u \hookrightarrow \ker(X \to \mathrm{coker} f)$ is an epimorphism, and also a monomorphism, thus it is an isomorphism.
A: I indeed had misunderstood the part of Kashira-Schapira you meant. While I don't know if the general result you say is true, I'll try to explain how I understood the fact that $\operatorname{im}f\cong\operatorname{ker}u$ when I studied abelian categories.

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