# Associative Product Bifunctor

Let $\mathcal{C}$ be a category with finite products. Under the assumption of the Axiom of Global Choice it is posible to define the functor $$\mathbf{Prod}\colon \mathcal{C} \times \mathcal{C}\hookrightarrow \mathcal{C}.$$ The functor $\mathbf{Prod}$ is such that

1. there exists $F_1\colon \mathrm{Ob}(\mathcal{C})\times\mathrm{Ob}(\mathcal{C})\to \mathrm{Hom}(\mathcal{C})$ and $F_2\colon \mathrm{Ob}(\mathcal{C})\times\mathrm{Ob}(\mathcal{C})\to \mathrm{Hom}(\mathcal{C})$;

2. the cone $(\mathbf{Prod}((X,Y)),F_1(X,Y),F_2(X,Y))$ is the product of $X$ and $Y$ for any $X,Y\in \mathrm{Ob}(\mathcal{C})$;

3. the arrow $\mathbf{Prod}((f,g))$ is the product arrow $f\times g$ with respect to products given in 2. for any arrows $X_1\stackrel{f}{\longrightarrow} X_2,Y_1\stackrel{g}{\longrightarrow} Y_2\in\mathrm{Hom}(\mathcal{C})$.

Is it possible to construct $\mathbf{Prod}$ that satisfy properties mentioned above and $$\mathbf{Prod}(\mathbf{Prod}(X,Y),Z)=\mathbf{Prod}(X,\mathbf{Prod}(Y,Z))$$ for any $X,Y,Z\in\mathrm{Ob}(\mathcal{C})$ and $$\mathbf{Prod}(\mathbf{Prod}(f,g),h)=\mathbf{Prod}(f,\mathbf{Prod}(g,h))$$ for any $f,g,h\in\mathrm{Hom}(\mathcal{C})$?

-
In this answer I give a construction of a strictly associative coproduct in the category of sets. If you are willing to change the underlying category $\mathcal{C}$, it is always possible to pass from a category $\mathcal{C}$ with a weakly associative product to an equivalent category $\mathcal{C}'$ with a strictly associative product. This is part of what the coherence theorem for monoidal categories tells us. – Zhen Lin Apr 1 '12 at 16:39
Thanks for the reference to monoidal categories. My question was motivated by some categories of algebraic structures. It seems like that the result of this paper [Turning monoidal categories into strict ones, ( New York Journal of Mathematics 7 (2001) 257-265)] implies the existence of the strict associative product bifunctor in my case. – Fedor Pakhomov Apr 2 '12 at 10:55