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Theorem. Suppose $f : X \rightarrow Y$ is monic. Then for all $g : \bar{X} \rightarrow Y$ there exists at most one $h : \bar{X} \rightarrow X$ such that $f \circ h = g$.

Question. Does the converse of the proposition hold? That is, is it true that for all $f : X \rightarrow Y$, if for all $g : \bar{X} \rightarrow Y$ there exists at most one $h : \bar{X} \rightarrow X$ such that $f \circ h = g$, then $f$ is monic?

And if not, what do we call morphisms $f : X \rightarrow Y$ such that for all $g : \bar{X} \rightarrow Y$ there exists at most one $h : \bar{X} \rightarrow X$ such that $f \circ h = g$?

Proof of the theorem. Suppose not. Then there exist two or more distinct such $h$, call them $h_0$ and $h_1$. Thus $f \circ h_0 = g$ and $f \circ h_1 = g.$ So $f \circ h_0 = f \circ h_1$. But since $f$ is monic, this implies $h_0 = h_1$. Contradiction.

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You quote the very definition of a monic morphism as a Theorem? Hilarious. –  Martin Brandenburg Apr 24 '13 at 20:22
    
Thanks, I do standup ;) –  goblin Apr 25 '13 at 0:23
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Sure. The proof is more or less the same. Suppose we have $h$ and $h'$ such that $f \circ h = f \circ h'$; then the hypothesis says $h = h'$. But this is exactly what it means to be monic.

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