$\DeclareMathOperator{\im}{Im}$
Let $\Phi: F_n\to F_n$ be an injective homomorphism between free groups with $n\geq 2$ and $\im{\Phi}$ having finite index. Prove that $\Phi$ is bijective, i.e. $\im{\Phi}$ has index 1.
Proof:
Let $\Phi: F_n \to F_n$ be an injective homomorphism between free groups of rank $n\geq 2$ with $\im{\Phi}$ having finite index in $F_n$. Since $\im{\Phi}$ is a subgroup of $F_n$ and by the Nielson-Schreier Theorem any subgroup of a free group is free, let $\im{\Phi}=F_m$ for some $m\geq 2$. Moreover, since the index of $F_m$ is finite therefore $F_m$ is free on precisely $k(n-1)+1$ generators for some positive integer $k$. Now, since $\Phi$ is injective then $\ker{\Phi}=1$, so by the First Isomorphism Theorem we will have
\begin{align*}
F_n / \ker{\Phi} = F_n / 1 &\cong \im{\Phi}=F_m\\
F_n &\cong F_m\\
\end{align*}
which means that $F_n$ and $F_m$ have the same rank, so $\Phi$ is an isomorphism, hence bijective. $\Box$
Is this correct?
As a background, I am allowed to know elementary group theory, as well as material from Chapter 1 of Algebraic Topology by Hatcher.
Best Answer
I cleaned up the reasoning in accordance with @user994373's comment above.
$\DeclareMathOperator{\im}{Im}$ Proof: Let $\Phi$ be an injective homomorphism from $F_n$ to itself for $n\geq 2$ and with $\im{\Phi}$ having finite index. Since $\Phi$ is injective, by the First Isomorphism Theorem we have $\ker{\Phi}=1$ so $F_n \cong \im{\Phi}$. Next, since $\im{\Phi}$ has finite index, by the Nielson-Schreier Theorem we have that $\im{\Phi}\leq F_n$ is a free subgroup of $F_n$, and in particular $\im{\Phi}=F_m$ where $m=k(n-1)+1$ for some positive integer $k$. Therefore we have $F_n \cong F_m$ and since isomorphic free groups have the same rank by the uniqueness property, therefore $n=m$, so $\im{\Phi}=F_n$ and $\Phi$ is a bijection. $\Box$