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According to Wikipedia, image of a morphism $\phi:X\rightarrow Y$ in a category is a monomorphism $i:I\rightarrow Y$ satisfying the following conditions:

  1. There is a morphism $\alpha:X\rightarrow I$ such that $i\circ\alpha=\phi$.
  2. If $j:J\rightarrow Y$ is a monomorphism and $\beta:X\rightarrow J$ is a morphism such that $\beta\circ j=\phi$, then there exists a unique morphism $\gamma:I\rightarrow J$ such that $\beta=\gamma\circ\alpha$ and $j\circ\gamma=i$.

It is easy to see that $\alpha$ is unique. Intuitively, such $\alpha$ should be an epimorphism, but I can't seem to show it. Is it true?

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2 Answers 2

up vote 5 down vote accepted

It's not true in absolute generality. For a counterexample, consider a category with five objects $A, B, C, D, E$, two morphisms $B \to C$ and $C \to D$ whose composite is $B \to D$, and two pairs of distinct parallel morphisms $A \rightrightarrows B$, which when composed with $B \to C$ give the same morphism $A \to C$, and $C \rightrightarrows E$, which when composed with $B \to C$ gives the same morphism $B \to E$. Clearly, $B \to C$ is not a monomorphism, $C \to D$ is a monomorphism and is the image of $B \to D$, but by construction $B \to C$ is not an epimorphism.

However, in a regular category, it is true that every morphism factors as a regular epimorphism followed by a monomorphism.

Edit. The previous version of the counterexample was not quite correct.

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The term "image" suggests that this concept is modeled on the image of a map, a morphism in the category of sets. In that case, $I$ can be any set equipotent with the image (in the conventional sense) of $\phi$, and $\alpha$ is generally neither unique, nor an epimorphism (a surjective map). For uniqueness, note that you can compose $\alpha$ with any permutation of $I$ to obtain another suitable $\alpha$ for a given $I$.

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I thought $i$ being monomorphism implies uniqueness of $\alpha$ ? –  ashpool Sep 2 '12 at 18:13

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