TITLE: The free monoid functor is fully faithful?
QUESTION [11 upvotes]: For every set $A$, there is a free monoid $A^*$ and a function $i_A : A \rightarrow A^*$, such that for all monoids $Z$ and functions $j : A \rightarrow Z$, there is a unique monoid morphism $j^* : A^* \rightarrow Z$, such that:
$$j^* \circ i_A = j$$
Consider the "free monoid functor" $F : \mathsf{Set} \rightarrow \mathsf{Mon}$. It maps a set $A$ to $A^*$ and a function $f : A\rightarrow B$ to $f^* : A^* \rightarrow B^*$.
I want to decide, whether $F$ is faithful, full or neither. I believe $F$ is full and faithful. In order to prove that, however, I apparently need, the fact that $i_A$ for every set $A$ is injective.
I want to use the UMP for that, but it is "weird" in the following sense: I have to somehow construct some arbitrary monoid structure on $A$, then we have:
$$id_A^* \circ i_A = id_A$$
So, in particular: $i_A$ is split mono and hence injective.
It is weird, that I have to make such a non-canonical choice, so I wonder:

Is there a way to prove more "canonically", that $i_A$ in this case, is injective?

Now, that we know that: 
Let $A,B$ be sets. Then the map $F_{A,B} : \mathsf{Set}(A,B) \rightarrow \mathsf{Mon}(A,B)$ has the inverse given by:
$$\tilde{f} \mapsto id^*_B\circ \tilde{f}\circ i_A$$
One confirms this, again, by using the UMP.
Therefore, $F$ is fully faithful.

Is my deduction correct?

(I also wonder, where I can look up simple facts like these, but I guess I'm going to ask another question for that later)

REPLY [5 votes]: Let $\mathbf C$ be a category and consider an adjunction $F\colon {\mathbf Set}  \leftrightarrow  \mathbf C\colon U$, with left adjoint $F$. Assume there exists some object $A$ in $\mathbf C$ with $|UA|\ge 2$. Then for all sets $S$ the unit $\eta \colon S\to UFS$ is injective. (The result you are interested in now follows by noting that monoids with more than one object exist.) 
Here is a sketch of the proof of the general claim. Fix some set $S$ and $f_1,f_2\colon *\to S$, two elements in $S$, and suppose that $\eta\circ f_1 = \eta \circ f_2$. The aim is to show that $f_1=f_2$. Computing the conjugate of $\eta \circ f_1 = \eta \circ f_2$ yields $Ff_1=Ff_2$. Now, because $|UA|\ge 2$ there exists a function $h\colon S\to UA$ and two functions $g_1,g_2\colon *\to UA$ with $g_i = h\circ f_i$. Computing the conjugate of $g_i = h\circ f_i$ yields $\bar h \circ Ffi$, and so $g_1=g_2$, and $f_1=f_2$ follows.