Suppose we have $u ; v ; w = id ; w$. Is it right to say that we also have $u ; v = id$?
I thought that this is like Leibniz, but the implication seems to be the wrong way round.
2026-05-05 13:45:24.1777988724
Inferring arrow equality
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1
No, there's no reason you can cancel $w$. If $w$ is a morphism to a terminal object, for instance, then the equation automatically holds for all possible choices.of $u,v$.