# Monos are exactly the “injective homomorphisms”

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 $x, y : 1 \rightarrow |M|$ ?

• "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 $M$. – SixWingedSeraph Feb 26 '16 at 3:03
• 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 $1$. – vkubicki Feb 26 '16 at 9:55

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 $f=g$.