Showing the rank of a group $H$ is precisely the maximal rank $\alpha$ of a sequence of groups. I'm working with an inverse limit of groups and I'm trying to prove some property. Let $F_n$ be the free group of rank $n$. We have a morphism from $f_n:F_n\to F_{n-1}$ that sends a word in $F_n$ to the word in $F_{n-1}$ with the instances of the $n$-th generator removed, it is surjective and by composing those morphism we have $\cdots\to F_n\to F_{n-1}\to \cdots \to F_2\to F_1$. We define $F$ as the inverse limit of this system, that is $F=\left\lbrace (w_n)_n\in\prod_i F_i \;\vert \;f_n(w_n)=w_{n-1}\right\rbrace$. My goal is to show that if $H$ is a finitely generated subgroup of $F$ then $H$ is free. I think I have the proof, there's just one detail that I can't quite figure out.
For each $n$ we have a projection $p_n:F\to F_n$, so let $H$ be a finitely generated subgroup of $F$ we define for each $n$ a group $H_n=p_n(H)<F_n$.
Since subgroups of free groups are free, each $H_n$ is free. I've shown that these free groups $H_n$ have a maximal rank $\alpha$, and that there is an $i_0\in \mathbb{N}$ such that for all $i\geq i_0$ the rank of $H_i$ is precisely $\alpha$.
So for $i\geq i_0$, the groups $H_i$ are the same (up to isomorphism since there's essentially one free group of each rank) and since finitely generated free group are Hopfian, the morphisms $f_n:H_n\to H_{n-1}$ (restricted to $H_n$) are isomorphisms. This allows us to show that $x\in H$ is solely determined by its image in $H_{i_0}$. So taking $B\subset H$ that is mapped to a free basis for $H_{i_0}$ by $p_{i_0}$, we can find a free basis for H and we're done. I've glossed over a bunch of details that are irrelevant to the question I'm about to ask but this is to give an idea of the context.
I'm fairly confident that I should be able to prove that the maximal rank $\alpha$ is precisely the rank of $H$, yet I can't seem to put my hand on why ? And since the rest of the proof sort of relies on it, it would help if I could prove it. Maybe something is wrong but I think it's supposed to work. At first I sort of hand waved it assuming filling in the details would be easy, but the more I think about it the less I know how to do it. I tried assuming $rank H > \alpha$ and then finding a contradiction to the fact that the morphism $f_i$ is an isomorphism (for $i\geq i_0$) but I didn't succeed. Not quite sure where to go from there.
EDIT: for clarification, the groups $H_n$'s rank is bounded by the rank of $H$ and its increasing with $n$ (or stationary) so it converges towards some $\alpha\in\mathbb{N}$. What I can't prove is that the rank of $H$ is not just an upper bound but the least upper bound, ie $rank\, H =\alpha$.
 A: The idea is that you have a restricted inverse sequence consisting of the $H_n$'s, its inverse limit is (identified with) the subgroup $H$ you are interested in, and $H$ is isomorphic to $H_n$ for $n \ge i_0$.
In more detail you have a commutative diagram of the form
$\require{AMScd}$
\begin{CD}
... H_{n} @>f_{n}  \mid H_{n}>> H_{n-1} @>f_{n-1} \mid H_{n-1}>> ... @>f_2 \mid H_2 >> H_1 \\
 @V \subset VV @V \subset VV @. @V \subset VV \\
... F_{n} @>f_{n}  >> F_{n-1} @>f_{n-1} >> ... @> f_2 >> F_1
\end{CD}
The bottom row is, of course, the given inverse sequence. The top row is an inverse sequence having the property that for sufficiently large $n$ the map $f_n \mid H_n : H_n \to H_{n-1}$ is an isomorphism for $n \ge i_0$.
It follows that the inverse limit $H$ of the top row is isomorphic to $H_n$ for $n \ge i_0$, and furthermore that the projection maps $H \mapsto H_n$ are isomorphisms for $n \ge i_0$.
It also follows, from the universal properties of inverse limits, that one has a commutative diagram
$$
\begin{CD}
H @>>> H_n\\
@V  V V @VV \subset V\\
F @>>> F_n
\end{CD}
$$
where the horizontal arrows are projection maps. Since the top and right arrows are known to be injective, it follows that $H \mapsto F$ is injective, and the image of this injection is the subgroup of $F$ that you want.
