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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 $N$?

1 Answers 1

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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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    Yeah, but how does he know that $g(a_1)*g(a_2)$ equals $g(a_1)\cdot_N g(a_2)$ and not, say, $g(a_3)\cdot_N g(a_5)$?2012-11-13
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    @drozzy: Because $g$ satisfies $g(a)=f(a)$ for all $a\in A$ (though you accidentally left that out the $=f(a)$ part of that) and is a monoid homomorphism, so necessarily $g(a_1\ast a_2)=g(a_1)\cdot_N g(a_2)$. The proof is by induction on the number of letters: $g(a)=f(a)$ for $a\in A$ takes care of single letters, and the homomorphism property gives you the induction step.2012-11-13
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    Oops, fixed. Reading your comment now. Thanks.2012-11-13
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    Ooh shoot. I forgot that $A^*$ was a monoid! I thought it was just a set! And I completely forgot the definition of the homomorphism for categories. Thanks for correcting me twice!2012-11-13
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    @drozzy: You’re welcome! (It happens; don’t worry too much about it.)2012-11-13
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    Thanks. I searched everywhere to find out the meaning of ⋅N. Now everything starts to make sense. I did not find the description in the book, either. Is that something everyone reads the book has already known?2016-09-18
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    @LiweiZ: It’s an interpretation that the reader is probably expected to be able to make fairly easily. It’s fairly common to subscript an operation symbol with the name of the algebraic object to indicate that it’s the operation in that particular object, and the line $$\bar f(a_1\ldots a_i)=f(a_1)\cdot_N\ldots\cdot_Nf(a_i)$$ adds enough context to make it clear that this has to be the intended interpretation.2016-09-18
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    @BrianM.Scott Thanks again for your lightning speed response. And thank you for making the detailed explanation for how the interpretation of the symbol is supposed to work.2016-09-18
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    @LiweiZ: You’re welcome.2016-09-18