Isomorphism of products in a category: Does it involve the axiom of choice?

Let $$(X_i), (Y_i)$$ be two families of objects indexed by the same index set $$I$$ in a category with products. It seems obvious that if $$X_i, Y_i$$ are isomorphic for all $$i$$, then also the products $$\prod_{i \in I} X_i$$ and $$\prod_{i \in I} Y_i$$ are isomorphic:

We know that for each $$i$$ there exists an isomorphism $$h_i : X_i \to Y_i$$. Then $$\prod_{i \in I} h_i : \prod_{i \in I} X_i \to \prod_{i \in I} Y_i$$ is an isomorphism.

But doesn't this involve the axiom of choice if $$I$$ is infinite? We have to make infinitely many choices to get a family $$(h_i)$$ of isomorphisms. Or can AC be avoided?

• Without AC, one could have a family of unordered two-element sets whose cartesian product is empty. "Unordered" means that there is no "first" or "second" element. Apr 23, 2021 at 16:31

Yes, the axiom of choice is being used. For example, suppose that there is a countable family of pairs $$P_n$$ such that $$\prod_{n\in\Bbb N}P_n=\varnothing$$. Let $$S_n=\{0,1\}$$ for all $$n$$, then there is an isomorphism $$h_n\colon P_n\to S_n$$, but still, $$\prod S_n$$ has $$2^{\aleph_0}$$ elements, while $$\prod P_n$$ has none.

The misleading part is that you write "there exists an isomorphism $$h_i$$", since it makes it seem that we chose one already. But we haven't, and we can't, not necessarily, without invoking the axiom of choice.

• @Karl: You are still working inside a set theory. When you say "choose $h_i$", you're not choosing them internally to the category, you're choosing them outside of the category, and you argue that you can combine them in a way that leads to a morphism or object in your category. If your category is not locally small, you might need more than just AC to choose morphisms (although depending on the exact set up, ZF can probably identify a set of morphisms between any two objects in a uniform way, so AC will still be more than enough for the argument). Apr 23, 2021 at 18:51
• @Karl: No, I'm referring to Scott's trick, which states that given any proper class, we can extract a canonical non-empty subset from it: simply look at those elements which have the least von Neumann rank in the class. Therefore, even if $\operatorname{Mor}(x_i,y_i)$ is a proper class for all $i\in I$, we can reduce them to sets uniformly so that we may apply the axiom of choice (for sets). Apr 23, 2021 at 19:32