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In Awodey, Category Theory, second edition, example 2.3 (page 30).

The example

In many categories of "structured sets" like monoids, the monos are exactly the "injective homomorphisms." More precisely, a homomorphism $h : M \rightarrow N$ of monoids is monic just if the underlying function $|h| : |M| \rightarrow |N|$ is monic, that is, injective by the foregoing. To prove this, suppose $h$ is monic and take two different "elements" $x, y : 1 \rightarrow |M|$ where $1 = \{\ast\}$ is any one-element set. By the UMP of the free monoid $M(1)$ there are distinct corresponding homomorphisms $\overline{x}, \overline{y} : M(1) \rightarrow M$ with distinct composites $h \circ \overline{x}, h \circ \overline{y} : M(1) \rightarrow M \rightarrow N$, since $h$ is monic. Thus, the corresponding "elements" $hx, hy : 1 \rightarrow N$ of $N$ are also distinct, again by the UMP of $M(1)$.

My question

To prove that $|h| : |M| \rightarrow |N|$ is monic, one must prove that for any set $C$ and functions $f, g : C \rightarrow |M|$, $(|h| \circ f = |h| \circ g) \Rightarrow (f = g)$. This is equivalent to proving by the contraposative that for any category $C$ and morphisms $f, g : C \rightarrow |M|$, $(f \neq g) \Rightarrow (|h| \circ f \neq |h| \circ g)$. Why is the proof in the mentioned example complete, as it is just proving this for $x, y : 1 \rightarrow |M|$ ?

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  • $\begingroup$ "just proving this for $x,y:1\to|M|$" means exactly that he is proving it for any two elements of $M$, since a map from $1$ to $M$ is the same thing as an element of $M$. $\endgroup$ – SixWingedSeraph Feb 26 '16 at 3:03
  • $\begingroup$ What I fail to understand is that he is supposed to prove it for any $f, g : C \rightarrow |M|$, and it seems to me he has only proven it for $f, g : 1 \rightarrow |M|$ so far. He has yet to prove it for any $C$ other than $1$. $\endgroup$ – vkubicki Feb 26 '16 at 9:55
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The missing step is just the immediately preceding result (Proposition 2.2) in Awodey, which says a map of sets is monic iff it is injective. The proof is very simple: in the notation of your question, given that $|h|$ is injective, if $|h|\circ f=|h|\circ g$, then $|h|(f(x))=|h|(g(x))$ for all $x\in C$, so $f(x)=g(x)$ for all $x\in C$ since $|h|$ is injective, so $f=g$.

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