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Let $\phi:A\rightarrow B$ be a morphism in a category, and $\phi':I\hookrightarrow B$ its image. Intuitively, $\phi$ should be an epimorphism if $\phi'$ is an epimorphism. But I have difficulty proving it using definitions of image and epimorphism (from Wikipedia). Is it true at all?

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up vote 4 down vote accepted

As usual with these things, it is not true in absolute generality. (Epimorphisms and images are very subtle creatures.)

Here is a counterexample. Let us have a category with five objects $A, B, C, D, E$, and non-identity arrows $A \rightrightarrows B$, $B \to C$, $C \to D$, $D \rightrightarrows E$, such that the composites $C \to D \rightrightarrows E$ are distinct while the pairs of composites $B \to C \to D \rightrightarrows E$ and $A \rightrightarrows B \to C$ are not distinct. By construction, $B \to C$ is not a monomorphism, $C \to D$ is a monomorphism and is the image of $B \to D$, while $C \to D$ is an epimorphism but $B \to D$ is not.

However, in any category in which (epimorphisms, monomorphisms) form an orthogonal factorisation system, your claim is true – for obvious reasons. For example, this is true in any topos and any abelian category.

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Thanks. Can you tell me why it is true in a balanced category? –  ashpool Sep 13 '12 at 17:09
    
Sorry, that's not quite right. I've replaced that with a correct statement. –  Zhen Lin Sep 14 '12 at 1:02
    
That's a relief to hear -- I spent the whole day trying to prove it, learning quite a lot of category theory in the process. Thanks! –  ashpool Sep 14 '12 at 2:02

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