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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 $u=\ker g \to X \to \mathrm{coker} f$.)

carciofo21
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    Is $A\to B\to C$ exact? Because there are examples of non-isomorphic $R$-modules $A$, $B$ such that $B\hookrightarrow A$ and $A\hookrightarrow B$, see here and here. –  Apr 19 '21 at 09:46
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    @Gae.S. S., if $A\to B\to C$ is exact and all the morphisms are monic, then $A=B=C=0$. – Gabriel Apr 19 '21 at 10:27
  • @Gae.S. This is inserted in the discussion of equivalent conditions for a sequence to be exact (so exactness should be irrelevant). – carciofo21 Apr 19 '21 at 10:27
  • @Gabriel Oh, yeah... That was silly. Teaches me trying to un-falsify statements. –  Apr 19 '21 at 10:29
  • Dear @carciofo21, I don't see where your affirmation is used in the reference cited. There, first of all, $X'\to X\to X''$ is a complex, not just an arbitrary collection of morphisms. Moreover, in no moment it is concluded that $X'\cong X$. – Gabriel Apr 19 '21 at 10:46
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    @Gabriel, yes, it is a complex, I've written that in the question. Anyway the book deduces an isomorphism $B=\mathrm{ker} u \simeq \mathrm{im} f=A$, as I mention in the last lines of my question, I don't claim that an isomorphism $X'\simeq X$ is inferred. – carciofo21 Apr 19 '21 at 11:35

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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.

Arnaud D.
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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.

If it interests you, you can find these notes here https://gabrielribeiro1707.wixsite.com/projects under "notes of everything".

Gabriel
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