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I got in an argument once about whether the statement:

"For all fields $F$ the category of finite-dimensional $F$-vector spaces is equivalent to the opposite of the category of $F$-matrices, briefly $\mathsf{FinVect}(F) \simeq \mathsf{Mat}^{\operatorname{op}}(F)$."

is provable in ZF. I argued against it saying something along the lines of "you would need to choose a basis in every vector space".

Is the statement provable in ZF? If not, how does it compare in strength to other consequences of the axiom of choice?

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    What exactly is the Mat category? Are matrices the objects or the arrows? – 2017-01-14
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    what is the category of F-matrices? – 2017-01-14
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    @Omnomnomnom The objects are all nonnegative integers. A morphism $m\to n$ is an $m\times n$ matrix. Composition is matrix multiplication and identities are identity matrices. – 2017-01-14
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    [Here](https://unapologetic.wordpress.com/2008/06/24/the-category-of-matrices-iv/) the statement is proved using the axiom of choice (careful the $\mathsf{Mat}$ there is the opposite of my $\mathsf{Mat}$). – 2017-01-14
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    Admittedly, I'm not sure where the axiom of choice comes into this game. – 2017-01-14
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    Sorry, one thing first: I'm really talking about the axiom of *global* choice; I forgot about the distinction. @AsafKaragila The statement that a fully faithful and essentially surjective on objects - functor can be "extended" to an equivalence of categories needs (some form of) global choice. I'm not sure whether it is equivalent right now, though, but it is definitely not provable in ZF+Classes. - I may need to change my question to reflect that I'm using classes... – 2017-01-14
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    In the section "Weak equivalence" [here](https://ncatlab.org/nlab/show/equivalence+of+categories) it says "A strict functor with a weak inverse is necessarily essentially surjective and fully faithful; the converse is equivalent to the axiom of choice.". I suppose global choice is meant, if one considers large categories. – 2017-01-14
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    What's your definition of finite-dimensional? I think you should be able to show (without choice) that $\mathsf{FinVec}$ is isomorphic to any subcategory of $\mathsf{Vec}$ generated by a (free) vector space of each finite dimension. Furthermore, I think that also without choice you should be able to show that such a subcategory is isomorphic to $\mathsf{Mat}^{op}$. So the issue is constructing such a subcategory without using choice, but why not simply take the full subcategory generated by the cartesian powers $F^n$? I think that only uses $\omega$-recursion or something, not choice. – 2017-01-14
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    (I'm not a set theorist, so I could be wrong on all three claims of course, but that's what I would try anyway) – 2017-01-14
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    @VladimirSotirov I have next to no doubt that the full subcategory generated by the $F^n$'s is equivalent to $\mathsf{Mat}^{\operatorname{op}}$ in **ZF** (I haven't proved it though). So in some sense my question possibly comes down too 'Is "the category of all finite vector spaces equivalent to the full subcategory on $F, F^2, \dots$" provable in **ZF**?'. – 2017-01-14
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    @VladimirSotirov A vector space is finite-dimensional if it has a finite basis. - The issue I'm thinking of is that I have not equipped the vector space with a finite basis, I merely assume there is one. – 2017-01-14
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    Stefan, maybe what bothers me is this problem: there is a proper class of finite dimensional vector spaces over a given field; but there is only a countable set of vector spaces which generate everything up to equivalence. And it seems to me that this is the issue you're grappling with maybe? – 2017-01-15
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    @Asaf Well, the size has something to do with the strength of the statement if it is indeed unprovable in ZF, but it shouldn't change the "fact" that some amount of choice is necessary. If this was about a *set* of countably many objects you would need countable choice and you need "more" choice if there are more objects. That is what I'm guessing. – 2017-01-15
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    I don't see why, though. If this is about the set of fields which are *exactly* $F^n$, then there is a canonical choice for a basis for each one. Every other finitely dimensional $F$-vector space is isomorphic to one of these. I haven't used choice just yet. So I'm not sure where choice kicks in. – 2017-01-15
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    (Moreover, if $V$ is a finite dimensional vector space, then I think that the categorical equivalence you're looking to prove will be uniformly induced by how you defined it on the "set of generators" (i.e. $F^n$'s) independently of the choice of basis for $V$. But that's just a hunch.) – 2017-01-15
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    @AsafKaragila I don't think that saves me from specifying *how* the other vector spaces are isomorphic to the $F^n$. – 2017-01-15
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    Yes, but my point is that up to conjugation, the choice of isomorphism shouldn't matter, and that this conjugation should, in theory, preserve all the arrows you're interested in. – 2017-01-15
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    @AsafKaragila I don't know exactly. I would probably need a very precise answer to my question. - While the choice doesn't matter, it seems that you have to make a choice regardless. – 2017-01-15
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    No, this is like saying that you need the axiom of choice to define the real numbers as equivalence classes of rational Cauchy sequences. Just because addition is defined by choosing representatives, does not mean that it depends on the choice of representatives. – 2017-01-15
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    @AsafKaragila Unfortunately, I'm not conviced this is what is going on here just by talking about it more or less superficially. Hence this question. – 2017-01-15
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    Stefan, and I don't understand the terms enough to give any proper answer. Hence my comments, probing into ideas and looking for feedback. – 2017-01-15
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    @Asaf Yes I understand. *I'm* not knowledgable enough to answer the question myself with the help of your comments, is what I'm saying. – 2017-01-15
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    Okay, maybe someone will come along and do the job for us! Let's wait. – 2017-01-15
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    @AsafKaragila Let $V$ be some $n$ dimensional vector space, and let $\phi:V\to F^n$ be an isomorphism. What is $S(\phi):n\to n$, where $S$ is the equivalence? How can you specify this image without choice – 2017-01-15
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    @user2520938: I guess that in my mind I was thinking about ordered bases, rather than just unordered bases. – 2017-01-15
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    @AsafKaragila: The problem here is that if you have weak inverses $L : \operatorname{FinVect} \leftrightarrow \operatorname{Mat}^\text{op} : R$, then the natural isomorphism $1 \to RL$ is a simultaneous choice of isomorphism $V \to R(n)$ for every $n$-dimensional $F$-vector space. If you furthermore pick a basis for each $R(n)$, you get a simultaneous choice of basis for all finite dimensional $F$-vector spaces. – 2017-01-16
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    @AsafKaragila: There are weak equivalences $R : \operatorname{Mat}^\text{op} \to \operatorname{FinVect} $, such as $R(n) = F^n$ and so forth as suggested. The problem is when you look to complete it into a strong equivalence. Alternatively, you could alternatively switch to anafunctors to be more forgiving about all these "up to isomorphism" problems. These categories should be anaequivalent even without a choice principle. – 2017-01-16

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Suppose $F$ is a field such that there is a total order on the underlying set of $F$. Then for any $n$ we have a lexicographical total order on $F^n$.

Suppose we had an equivalence $S: \mathsf{FinVect}(F) \to \mathsf{Mat}^{\operatorname{op}}(F)$.

Let $\mathcal{A}$ be a set of two-element sets.

For any $A\in\mathcal{A}$, $F^A$ is a $2$-dimensional vector space over $F$, as is $F^2$, and so $S(F^A)=S(F^2)$. The identity map $S(F^A)\to S(F^2)$ is $S(\varphi)$ for a unique isomorphism $\varphi:F^A\to F^2$.

Let $\{v_1,v_2\}$ be the natural basis of $F^A$, so there is a natural bijection $A\to\{v_1,v_2\}$, and choose $v_1$ or $v_2$ depending on whether or not $\varphi(v_1)>\varphi(v_2)$ in the lexicographical order.

This gives a choice function for $\mathcal{A}$. But choice for two-element sets is independent of $\text{ZF}$.