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I'm asking about one type of proof of epic $\implies$ surjective like Clive Newstead's answer in Epic implies Surjective or Thomas Andrew's answer in Morphism epimorphism if and only if surjective. For reference, here is the definition of epic that I'll be using (the same definition as in the first linked post):

A function $f \colon A \to B$ is epic if for all $Z$ and all $\alpha, \alpha^{\prime} \colon B \to Z$, $\alpha \circ f = \alpha^{\prime} \circ f \implies \alpha = \alpha^{\prime}$.

Now, I can understand why the proofs work in sets $Z$ with at least two distinct elements. But if $Z$ has one element, no matter the set $B$ we have $\alpha = \alpha^{\prime}$ and the implication $\alpha \circ f = \alpha^{\prime} \circ f \implies \alpha = \alpha^{\prime}$ holds trivially. Then the choice of $Z$ necessarily means that $f$ is epic, but $f$ need not be surjective: for example we can have $B = \{0, 1\}$, $Z = \{1\}$, and $\forall x \in A$, $f(x) = 0$. So why do the proofs in the linked posts assume that $Z$ has at least two distinct elements?

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The definition requires the implication to hold for every $Z$. It is exactly because of your observation (i.e. the implication always hold when $|Z|=1$) that we only need to prove the implication holds for $|Z|>1$ when proving $f$ is epic.

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  • $\begingroup$ Ah, of course I misread the definition. Thanks for the help! $\endgroup$
    – sanguine
    Jul 24, 2021 at 22:20

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