Question on Universal Mapping Properties

Question

In Steve Awodey's book Category Theory, he has some sentence that sounds confusing to me on page $57,$ he mentioned that

The universal mapping properties ($UMP$s) of $M(A),M(B),A+B,$ and $M(A+B)$ then imply that the last of these has the required UMP of $M(A)+M(B).$

Here $M(X)$ is the set of a monoid, $X$ is the generating set.

The issue about my understanding is what he referred to the last? And how universal mapping property plays an role in proving that the universal mapping properties of $M(A)+M(B).$

From my view of understanding, the sentence means that if we have the UMPs of $M(A),M(B), A+B,M(A+B),$ we will have UMP of $M(A)+M(B).$ But how can it follow? I want to see some commutative diagram arguments on it.

Edit:

Moreover, later Awodey mentioned that coproduct $M(A)+M(B)$ is not the coproduct of the underlying set. But I am curious about the reason why?

Answer 1

If you connect universal properties to representability, you can use the following calculation which immediately generalizes and is the proof that left adjoints preserve coproducts (and, more generally, colimits via essentially the same argument): $$\begin{align}\mathbf{Mon}(M(A+B),N) &\cong \mathbf{Set}(A+B,U(N))\\ &\cong \mathbf{Set}(A,U(N))\times\mathbf{Set}(B,U(N))\\ &\cong \mathbf{Mon}(M(A),N)\times\mathbf{Mon}(M(B),N)\\ &\cong \mathbf{Mon}(M(A)+M(B),N)\end{align}$$ This is natural in $N$ and by Yoneda this means $M(A+B)\cong M(A)+M(B)$. $U(N)$ stands for the underlying set of the monoid $N$.

The first isomorphism corresponds to the UMP for $M$ and the second for the UMP for $+$. Then we use the UMP for $M$ twice and finally UMP for $+$.

If Awodey has introduced representability already, then you should show that the above (natural in $N$) isomorphisms are indeed equivalent to the relevant UMPs. If not, you can use the above as a guide for the argument, or you can just skip to where Awodey discusses representability because it really does make many categorical proofs very straightforward.

Answer 2

The other answer is really the best way to see this, but from scratch, we can copy Awodey's diagram and try to parse it. There is a bit of a cheat here because if the diagram is in $\text{Set}$, then the second row really should be $UM(A)\rightarrow UM(A+B)\leftarrow UM(B)$ where $U:\text{Mon}\to \text{Set}$ is the forgetful functor. This doesn't really solve the problem, though, because the top arrows are in $\text{Mon}.$ So it's best to see this diagram as two diagrams, one in $\text{Set}$ and one in $\text{Mon}$. I will assume you know that $M$ and $U$ are functors that satisfy the universal property that $[MA,N]\cong [A,UN]$ via $f\mapsto Uf\circ \eta_A$. Then,

$1).\ $ In $\text{Set},\ $ the bottom row constitutes a coproduct diagram, in which $A+B$ is the coprduct and $i_A:A\to A+B$ and $i_B:B\to A+B$ are the corresponding morphisms.

$2).\ $ In $\text{Mon}$, we have morphisms $Mi_A:MA\to M(A+B)$ and $Mi_B:MB\to M(A+B).$ The claim is that $M(A+B),$ together with these morphisms, is a coproduct. To show this, we take any object $N$ and morphisms $f:M(A)\to N$ and $g:M(B)\to N$ and produce a unique $\phi: M(A+B)\to N$ such that $\phi\circ Mi_A=f$ and $\phi\circ Mi_B=g.$

To do this, note:

$3).\ $ the UMP of the coproduct (in $\text{Set}$) tells us that there is a unique morphism $\psi:A+B\to UN$ such that $Uf\circ \eta_A=\psi\circ i_A$ and $Ug\circ \eta_B=\psi\circ i_B$, so now the UMP of $M$ gives us a unique arrow $\phi:M(A+B)\to N\ (\text{in Mon})$ such that $U\phi\circ \eta_{A+B}=\psi.$

$4).\ $ But (again by the UMP of the coproduct $A+B,)$ we have $UMi_A\circ \eta_A=\eta_{A+B}\circ i_A$ so $U(\phi\circ Mi_A)\circ \eta_A=U\phi\circ UMi_A\circ \eta_A=U\phi\circ \eta_{A+B}\circ i_A=\psi\circ i_A=Uf\circ \eta_A$ and now the UMP of $M$ tells us that $\phi\circ Mi_A=f$, as required. The same type of argument shows that $\phi\circ Mi_B=g.$