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Assume we have two categories $\cal A,B$ and four functors $F,G:\cal A\to B$, $D:\cal A\to A$, $E:\cal B\to B$.

In order to show an equality like $$E\circ F=G\circ D\quad (1)$$ I have to prove that this holds for all Objects $\textit{and}$ morphisms in $\cal A$. Or is there more to do?

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  • $\begingroup$ No need to do anything else, it's enough to show such an equation for all morphisms. (Perhaps the equations become isomorphisms..) $\endgroup$ – Berci Sep 7 '18 at 22:49
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Yes, those two functors $(E\circ F)$ and $(G \circ D)$ will be equal iff they are equal "pointwise" on all objects and all morphisms.

As a shortcut, note that if you prove that they're equal on all morphisms, then you've proven that they're equal on all objects as well — otherwise, the identity morphisms wouldn't be equal:

$$(E\circ F)(X) = (G\circ D)(X) = Y \\\Longleftrightarrow (E\circ F)(id_X) =(G\circ D)(id_X) = id_Y$$

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