# Extending a product in a category

Consider category $\mathbf{Top}$ of topological spaces and category $\mathbf{Fcd}$ of endofuncoids. (To answer my question all you need to know about $\mathbf{Fcd}$ is that $\mathbf{Top}$ is its full subcategory, if you want more details read the draft of the second volume of my book.)

Someone has said me the following about the attempt to extend Tychonoff product from $\mathbf{Top}$ to $\mathbf{Fcd}$:

Let $i$ be the inclusion functor from $\mathbf{Top}$ to $\mathbf{Fcd}$.

If $i$ has left adjoint: If $r$ is left adjoint to $i$, we have $\operatorname{Hom}(A, i(X\times Y)) = \operatorname{Hom}(r(A), X\times Y) = \operatorname{Hom}(r(A), X)\times \operatorname{Hom}(r(A), Y) = \operatorname{Hom}(A, i(X))\times \operatorname{Hom}(A, i(Y)).$

If also the left adjoint is full and faithful: $\operatorname{Hom}(A, i(r(X)\times r(Y))) = \operatorname{Hom}(r(A), r(X)\times r(Y)) = \operatorname{Hom}(r(A), r(X))\times \operatorname{Hom}(r(A), r(Y)) = \operatorname{Hom}(A, X)\times \operatorname{Hom}(A, Y).$

I doubt the formula $\operatorname{Hom}(r(A), X\times Y) = \operatorname{Hom}(r(A), X)\times \operatorname{Hom}(r(A), Y)$. Isn't it requiring that $\mathbf{Top}$ is monoidal closed? I think it isn't.

Also I doubt the formula $\operatorname{Hom}(r(A), r(X)\times r(Y)) = \operatorname{Hom}(r(A), r(X))\times \operatorname{Hom}(r(A), r(Y))$, because it is yet unknown whether $\mathbf{Fcd}$ is monoidal closed.

I also don't understand why $\operatorname{Hom}(r(A), r(X))\times \operatorname{Hom}(r(A), r(Y)) = \operatorname{Hom}(A, X)\times \operatorname{Hom}(A, Y)$.

Was the guy who wrote these formulas wrong? Or is it my bad understanding of his formulas?

1. $\hom(r(A),X\times Y)\cong \hom(r(A),X)\times\hom(r(A),Y)$ follows from the very definition of $X\times Y$ and doesn't require anything else.
The same holds to the next question with $r(X)$ and $r(Y)$.
2. The left adjoint - that you called $r$ (I guess because it is the reflection functor) - was assumed to be full and faithful here, i.e. bijective on homsets.
(The equality was meant isomorphic (in $\bf{Set}$).)