I'm trying to prove directly that $\text{Hom}(A, \prod_i C_i) \simeq \prod_i \text{Hom}(A, C_i)$ whenever $\prod_{i\in I} C_i$ is a product of a family of objects $C_i$ in a category $B$, where $A$ is any other object of $B$.

I want to do this using the universal product property of product. So far I've got that there is a unique map $u : \text{Hom}(A, \prod_i C_i) \to \prod_i \text{Hom} (A, C_i)$ such that $p_i \circ u = q_i$ where $p_i : \prod_i \text{Hom}(A, C_i) \to \text{Hom}(A, C_i)$ is the projection in the definition of product and $q_i : \text{Hom}(A, \prod_i C_i) \to \text{Hom}(A, C_i)$ is any family of maps to the components. The reason there is a unique map is by definition of the product $\prod_ i \text{Hom}(A, C_i)$ and its universal property.

I wanted to show that a universal arrow exists from $\prod_i \text{Hom}(A, C_i) \to \text{Hom}(A, \prod_i C_i)$ so that we'd have:

$$ p_i = q_i \circ v, !v \\ q_i = p_i \circ u, !u $$

so that $p_i = p_i \circ (u\circ v)$ and by the universal property self-applied to the product yields $u \circ v = \text{id}$. But I'm not seeing any obvious choice for $v$. That's where I need help.


1 Answer 1


For that other direction, you'll have to use the specifities of $\mathbf{Set}$ : to define a map $\prod_{i\in I} \hom(A,C_i) \to \hom(A,\prod_{i\in I}C_i)$ it suffices to define it on each and every element of the LHS.

So you can take $(f_i)_{i\in I}\in \prod_{i\in I}\hom(A,C_i)$, this means that for each $i\in I, f_i : A\to C_i$. Then this yields $A\to \prod_{i\in I}C_i$ by the universal property, etc.etc. I'll leave the rest to you.

(note that you can't use a universal property to define the map in the other direction, because you can map into products, not easily out ot them - that's where the specificities of $\mathbf{Set}$ come in -, and the RHS $\hom(A,\prod_{i\in I}C_i)$ has a priori no universal property - we're trying to prove that it has one !-)

  • $\begingroup$ Thank you for leaving it as a hint and not just giving away the answer. Ad victorem spoiles! $\endgroup$ Feb 8, 2020 at 16:25

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