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

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    It's so simple?2011-09-09
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    @Hans: It couldn't really be anything else, since in general there's no reason for there to *be* any other morphisms into $A$.2011-09-09
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    and the converse?2014-04-09
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    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](http://ncatlab.org/nlab/show/generalized+element) 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.2015-07-10
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    The article is "[Doing without Diagrams](http://www.maths.ed.ac.uk/~tl/elements.pdf) by Tom Leinster.2015-07-10
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    The article by "[Doing Without Diagrams](http://www.maths.ed.ac.uk/~tl/elements.pdf)" by Tom Leinster on the other hand seems to confirm the above answer...2015-07-10