Show that for each integer $n ≥ 2$, $F_n$ is a finite index subgroup of $F_2$, where $F_n$ is the free group on $n$ generators. I'm trying to answer a problem which asks me to show (using covering spaces) that for each integer $n ≥ 2$, $F_n$ is a finite index
subgroup of $F_2$, where $F_n$ is the free group on $n$ generators.
I can see that if I can find some covering space with fundamental group $F_n$ of the wedge $S^1 \vee S^1$ then this will give me the existence of the subgroup, and I know that this will be of finite index if this covering map sends finitely many points to the vertex joining the two circles in $S^1 \vee S^1$.
The problem I'm having is I'm struggling to find such a covering space - the only space I know of with fundamental group $F_n$ is the wedge of $n$ circles, but I can't see any covering map from that to $S^1 \vee S^1$ so I assume I must need another.
If anyone could suggest such a covering space (and if so how I can show it has fundamental group $F_n$) I'd really appreciate the help!
 A: You might be interested in reading the first few pages of Hatcher's section on covering spaces (starts page 56, 65 in pdf). I'll be taking the pictures from there.
let the wedge $S^1 \vee S^1$ be labeled as follows:

And consider the space of $n$ circles connected together $\bigcirc\!\!\bigcirc\cdots \bigcirc\!\!\bigcirc$ as in the post linked in the comments. This will be our covering space.
We're going to label this space as follows:

(This is a particular case, but you get the idea)
Hopefully it's clear that if I send each labeled edges (preserving orientation) to their corresponding edges in $S^1 \vee S^1$, we get a covering map. It remains to show that this space has fundamental group of $F_n$.
To do this, imagine taking the bottom left $b$ edge and gradually contracting it until until it disappears (the two endpoint are now one). The upper left $b$ is now a circle, so we have two circles (left $a$, upper left $b$) wedged with a figure eight. This process doesn't change the fundamental group (fill in the details). 
Do the same thing with the bottom right $a$. We now have a $4$-rose! Because the process doesn't change the fundamental group, this means the space we started with has fundamental group $F_4$. 
It seems like you know how to complete the rest of the proof about finite index subgroups.
