I started brushing up on my understanding of adjunctions and came across this well-known fact (rephrased in my own words):
Let $\mathbb{C}$ be a category, and let $\Delta:\mathbb{C} \to \mathbb{C} \times \mathbb{C}$ be the functor mapping objects $C$ of $\mathbb{C}$ to objects $(C, C)$ in $\mathbb{C} \times \mathbb{C}$.
$\mathbb{C}$ has binary products if and only if $\Delta$ has a right adjoint.
The case where $\mathbb{C} = \mathsf{Set}$ is discussed in the introduction to Categories for the Working Mathematician and makes sense to me.
The more general fact is proven in Awodey's Category Theory and also mentioned on Wikipedia and other Stackexchange questions.
However, I'm missing something to fully understand the available proof(s). I can see how $\mathbb{C}$ having products implies the existence of a right adjoint (namely the product functor). I do not understand though how the existence of a right adjoint implies that $\mathbb{C}$ has binary products.
To rule out any conceptual misunderstandings, let me clarify what I think it is that needs to be shown:
For all $X, Y \in \mathrm{obj}(\mathbb{C})$ there exists an object $X \times Y \in \mathrm{obj}(\mathbb{C})$ together with morphisms $X \times Y \xrightarrow{\pi_1} X$ and $X \times Y \xrightarrow{\pi_2} Y$ such that for all $C \in \mathrm{obj}(\mathbb{C})$ and any two morphisms $C \xrightarrow{f} X$ and $C \xrightarrow{g} Y$ there exists a unique morphism $C \xrightarrow{\langle f, g \rangle} X \times Y$ such that $f = \pi_1 \circ \langle f, g \rangle$ and $g = \pi_2 \circ \langle f, g \rangle$.
This is how I would start:
Suppose there exists $R : \mathbb{C} \times \mathbb{C} \rightarrow \mathbb{C}$ with $\Delta \dashv R$. Then for all $C \in \mathrm{obj}(\mathbb{C})$ and all $(X, Y) \in \mathrm{obj}(\mathbb{C} \times \mathbb{C})$, we have that $\mathrm{hom}(\Delta(C), (X, Y)) \cong \mathrm{hom}(C, R(X, Y))$.
It follows that we can associate each morphism $(C, C) \xrightarrow{(f, g)} (X, Y)$ (i.e. each pair of morphisms $C \xrightarrow{f} X$ and $C \xrightarrow{g} Y$) with a unique morphism $C \xrightarrow{\langle f, g \rangle} R(X,Y)$.
This is where I get stuck (and where all other proofs that I've found end).
How do we know that there exist morphisms $R(X, Y) \xrightarrow{\pi_1} X$ and $R(X, Y) \xrightarrow{\pi_2} Y$ such that $f = \pi_1 \circ \langle f, g \rangle$ and $g = \pi_2 \circ \langle f, g \rangle$?
Best Answer
One thing to think about is, have you used all of the things which you assumed? One of the requirements of an adjunction, is that maps $$ \alpha_{c,(x,y)}\colon \mathrm{Hom}(\Delta c, (x,y))\xrightarrow{\cong} \mathrm{Hom}(c, R(x,y)) $$ assemble into a natural isomorphism. For an object $c$ with maps $f\colon c\to x, g\colon c \to y$, we get a unique map $\langle f, g\rangle \colon c\to R(x,y)$ by the adjunction. Using this morphism, we obtain the following commutative sqaure: $$ \require{AMScd} \begin{CD} \mathrm{Hom}(\Delta R(x,y), (x,y)) @>{\cong}>> \mathrm{Hom}(R(x,y), R(x,y))\\ @V{(\Delta\langle f,g\rangle)^*}VV @V{\langle f,g\rangle^*}VV \\ \mathrm{Hom}(\Delta c, (x,y)) @>{\cong}>> \mathrm{Hom}(c, R(x,y)) \end{CD} $$ Starting with $\mathrm{id}_{R(x,y)}$ on the top right, the following elements correspond: $$ \require{AMScd} \begin{CD} (\pi_1,\pi_2)@>{\alpha}>> \mathrm{id}_{R(x,y)}\\ @V{(\Delta\langle f,g\rangle)^*}VV @V{\langle f,g\rangle^*}VV \\ (f,g) @>{\alpha}>>\langle f, g \rangle \end{CD} $$ where $\pi_1,\pi_2$ are the names of the maps from $R(x,y)$ to $x$ and $y$ respectively, which show up under the horizontal bijection. The $\pi_i$'s are actually independent of $c$, which is what we want. The fact that $\Delta(\langle f,g\rangle)^*(\pi_1,\pi_2)= (f,g)$ precisely means that $\pi_1\circ \langle f,g\rangle = f$ and $\pi_2\circ \langle f,g\rangle = g$.