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I understand why in the category of sets two parallel morphisms $f, g: A \rightarrow B$ are identical iff for each element $x: 1 \rightarrow A$ it holds that $f\circ x = g \circ x$.

Awodey on p. 36 of Category Theory asks (as an exercise), why in any category two parallel morphisms $f, g: A \rightarrow B$ are identical iff for each generalized element $x: X \rightarrow A$ it holds that $f\circ x = g \circ x$.

Could someone please give me a hint how to prove this?

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up vote 9 down vote accepted

Just let $X=A$ and $x$ be the identity morphism.

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It's so simple? – Hans Stricker Sep 9 '11 at 10:09
@Hans: It couldn't really be anything else, since in general there's no reason for there to be any other morphisms into $A$. – Chris Eagle Sep 9 '11 at 10:11
and the converse? – user42912 Apr 9 '14 at 9:23
The hint in the book was that this had to be the case for "each generalized element x:X→A" . If only one relation had sufficed then the question would have been phrased differently. NLab seems to suggest that this is related to the Yoneda Embedding, but that is beyond my ability to understand. An answer that explains this intuitively would be nice. – Henry Story Jul 10 '15 at 18:13
The article is "Doing without Diagrams by Tom Leinster. – Henry Story Jul 10 '15 at 20:26

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