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While I'm reading the Algebra: Chapter 0 by Paolo Aluffi, I've encountered the following paragraph where he discuss that the isomorphism on categories is too restrictive:

When are two categories ‘equivalent’?
Essentially every notion of ‘isomorphism’ encountered so far has boiled down to ‘structure-preserving bijection’, and the reader may well expect that the natural notion identifying two categories should be drawn from the same model: a functor matching objects of one category precisely with objects of another, preserving the structure of morphisms.

One could certainly introduce such a notion, but it would be exceedingly restrictive. Recall that solutions to universal problems, that is, just about every construction we have run across, are only defined up to isomorphism. Requiring solutions in one context to match exactly with solutions in another would be problematic. The structure of an object in a category is adequately carried by its isomorphism class”, and a natural notion of ‘equivalence’ of categories should aim at matching isomorphism classes, rather than individual objects. The morphisms are a more essential piece of information; the quality of a functor is first of all measured on how it acts on morphisms.

My problem lies in the part

Requiring solutions in one context to match exactly with solutions in another would be problematic.

while I can understand that the notion of functor isomorphism seems too restrictive on categories, I do want to further convince myself by an actual example. Is there an example that can justify the sentence "requiring solutions in one context to match exactly with solutions in another would be problematic"?

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    $\begingroup$ The universal solution to some problem in $\mathbf{Vect}_k$ is often a vector space of dimension $n$. It really doesn't matter how you want to call your basis elements, be they arrows in Euclidean space or formal linear combinations of vegetables. However, strictly speaking, those spaces are not the same but "just" isomorphic - which of course is usually more than enough. The same applies to basically every category. $\endgroup$
    – Qi Zhu
    Feb 27 at 15:22
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Consider the category of sets and the operation of the cartesian product of sets. It is rather straightforward to prove that the cartesian product operation is associative up to canonical isomorphisms. This is simply a matter of moving parenthesis in the syntactic description of the cartesian products as tuples but also follows from the universal property of products. Now, you define the notion of cardinality of a set as follows. Let $C$ be the category of sets and bijections and let $Card$ be a skeleton of it. There is an inclusion $Card \to C$ which is an equivalence but not an isomorphism. Now we wish to define, say, the product of cardinalities. We do so by $Card \times Card \to C\times C \to C \to Card$ using the injection followed by the cartesian product followed by the quasi-inverse of the inclusion. Now we phrase the problem: is the product of cardinalities associative? To answer that we translate the problem back to $C$ where we do not get an equality between the sets representing the problem but we do get an isomorphism. That isomorphism tells us that back in $Card$ we do have an equality. So the solutions in $C$ do not match exactly with the solutions in $Card$. But the equivalence is sufficient.

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  • $\begingroup$ Excellent answer :) $\endgroup$
    – macton
    Feb 27 at 5:47

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