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In the following proof $A^*$ here is a Kleele closure and $*$ is a product of $a$'s or "concatanation":

Proposition 1.9. $A^*$ has the UMP of the free monoid on A.
Proof. Given $f:A\to|N|$, define $\bar{f}:A^* \to N$ by
$\bar{f}(-) = u_N$, the unit of $N$
$\bar{f}(a_1 ... a_i) = f(a_1) \cdot_N ... \cdot_N f(a_i)$

Then $\bar{f}$ is clearly a homomorphism with
$\bar{f}(a) = f(a)$ for all $a \in A$

If $g:A^* \to N$ also satisfies $g(a)=f(a)$ for all $a\in A$, then for all $a_1 ... a_i \in A^*$:

$ g(a_1 ... a_i) = g(a_1 * ... * a_i)) \\ =g(a_1) \cdot_N ... \cdot_N g(a_i)$

How does the author get from $*$ to $\cdot_N$?
I mean how does the author know that $g$ also has a $\cdot_N$ operator in $N$?

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The operation $\cdot_N$ is the monoid operation in $N$; it has nothing to do specifically with $f$ or $g$. In that last line the objects $g(a_1),\dots,g(a_i)$ are elements of $N$, simply because $N$ is the codomain of $g$, so they can be combined using the operation $\cdot_N$ of $N$, just like any other elements of $N$.

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    @LiweiZ: You’re welcome.2016-09-18