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It is a well known result that given two isomorphic free groups, their freely generating sets have the same cardinality, which is then called the rank of a free group.

I am well aware of the usual proof that the rank of a free group is well defined, by abelianization and (if this is not reduced far enough for you) tensoring with $\mathbb{Q}$ to obtain a vector space. However I was wondering, whether there is a more abstract way to prove this result relying solely on the free functor satisfying some categorical properties.

Taking the perspective of category theory one might be tempted to rephrase this fact as follows:

The free functor $F: \underline{Set} \rightarrow \underline{\smash{Grp}}$ creates isomorphisms, in the sense that given an isomorphism $\phi: F(G) \rightarrow F(H)$ there is a morphism $f:G \rightarrow H$ with $F(f) = \phi$ and $f$ is an isomorphism.

EDIT: I was rightfully corrected that it is wrong that any isomorphism between free groups is in the range of the free functor. Nevertheless my question about reflecting the property of being isomorphic is still open.

I think I managed to show that the free functor reflects isomorphisms, ie. if $F(f)$ is an isomorphism, so is $f$. I showed that the free functor is faithful and thus reflects monos and epis. But as monos and epis in $\underline{Set}$ and $\underline{\smash{Grp}}$ split, it also reflects isos.

Now the problem is just, whether each isomorphism between free groups is in the range of the free functor. Clearly the free functor is not full (take $\mathbb{Z} \rightarrow F_2, 1 \mapsto ab$). However I could not think of a similiar counterexample which also is an isomorphism.

So my question is: Is my reformulation correct and if so, is there a categorical proof, ie. does the free functor satisfy some categorical criterion which makes it create isomorphisms?

Thank you all for your time!

Jonas Linssen
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  • The question "whether each isomorphism between free groups is in the range of the free functor" has a fairly simple negative answer. For example, there is a unique automorphism of the free group on one generator $x$ that sends $x$ to $x^{-1}$, but it is clearly not in the image of the free functor. Similarly, with two generators $x$ and $y$, you can send $x$ and $y$ to $x$ and $xy$, for example. – darij grinberg Jun 07 '19 at 16:45
  • Possible duplicate of free groups: $F_X\cong F_Y\Rightarrow|X|=|Y|$ – darij grinberg Jun 07 '19 at 16:46
  • The other question is answered at https://math.stackexchange.com/questions/35229/free-groups-f-x-cong-f-y-rightarrowx-y and at https://math.stackexchange.com/questions/110459/is-there-a-simple-proof-of-the-fact-that-if-free-groups-fs-and-fs-are-i . (Abelianization is a fairly functorial construction, isn't it?) – darij grinberg Jun 07 '19 at 16:46
  • @darijgrinberg Thanks for the counterexamples. I might adapt my question in such a matter that the first one is not a counterexample anymore, the second one is a bit harder to tackle though...

    I do not think the other two posts answer my question completely, as I am asking for categorical properties which make the free functor act like he does. It might however be that there is no such property...

     – Jonas Linssen Jun 07 '19 at 17:05
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    The answers to the questions linked above still depend intrinsically on tensoring with $\mathbb Q$ or perhaps with some other field, and then applying well-definedness of dimension of a vector space over whatever field you are using. Forbidding one from using that construction seems to me to be the point of this question, and it is an interesting point. – Lee Mosher Jun 07 '19 at 18:43

1 Answers1

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I don't know about a categorical proof, but there is a topological proof based on the method of Stallings fold paths. Here is a link to a paper which carries out this proof in an algorithmic fashion, starting from a given isomorphism $F(A) \mapsto F(B)$. The process is explicit enough that perhaps it can be turned into a categorical proof.

Lee Mosher
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