Any countable free group is embeddable into a free group of rank $2$ I have found this proof of the fact that any countable free group is embedable in a free group of rank $2$ (see the last page, Proposition 2). But isn't this proof incorrect?
First off, it says $w=b^{-i_1}a^{\epsilon_1}b^{i_1}\dots$. Shouldnt it be $w=b^{-\epsilon_1 i_1}a^{\epsilon_1}b^{\epsilon_1  i_1}\dots$? Or are they somehow equivalent.
Second, it says $a^{\epsilon_j}$ and $a^{\epsilon_{j+1}}$ are present in the literal and so $w$ cannot collapse to $1$. But that is not true...it even says that $i_j$ may equal $i_{j+1}$, and if so then we must collapse $a^{\epsilon_j}a^{\epsilon_{j+1}}$ as $a^{\epsilon_j+\epsilon_{j+1}}$, and so clearly these two literals are not present in $w$ in this case, and so how then do we know that after perhaps collapsing it some more we do not get the exponent of this to equal $0$?

Note that $$x_{i_1}^{\epsilon_1}=b^{-i_1}a^{\epsilon_1}b^{i_1}$$
Thee is no collapsing when going from $a^{\epsilon_j}a^{\epsilon_{j+1}}$ to $a^{\epsilon_j+\epsilon_{j+1}}$. Note that we speak about words over the alphabet $\{a,b,a^{-1},b^{-1}\}$ and that $a^n$ is just a notational shorthand for $\underbrace{aaa\ldots a}_n$ if $n>0$ or $\underbrace{a^{-1}a^{-1}a^{-1}\ldots a^{-1}}_{|n|}$ if $n<0$ or the empty word if $n=0$; at any rate $a^n$ stands for $|n|$ specific letters of our alphabet. Collapsing would mean that the total number of letters changes, but in $a^{\epsilon_1}a^{\epsilon_2}$ we have $|\epsilon_1|+|\epsilon_2|$ letters and in $a^{\epsilon_1+\epsilon_2}$ we have $|\epsilon_1+\epsilon_2|$ letters. We have $|\epsilon_1+\epsilon_2|=|\epsilon_1|+|\epsilon_2|$ because either $\epsilon_1=\epsilon_2=+1$ or $\epsilon_1=\epsilon_2=-1$. (In other words, neither $aa$ nor $a^{-1}a^{-1}$ allows collapsing).
