I'm studying the proof that $S_n = \langle s_1,...,s_{n - 1} \mid s^2_i = 1, (s_is_{i + 1})^3, s_is_j = s_js_i$ for $|i - j| > 1 \rangle$. The key part of the proof is that the subgroup of $\Gamma_{n + 1} = \langle s_1,...,s_n \mid s^2_i = 1, (s_is_{i + 1})^3, s_is_j = s_js_i$ for $|i - j| > 1 \rangle$ generates by $s_1,...,s_{n - 1}$ may be realized as a subgroup of $\Gamma_n = \langle s_1,...,s_{n - 1} \mid s^2_i = 1, (s_is_{i + 1})^3, s_is_j = s_js_i$ for $|i - j| > 1 \rangle$.
To justify it formally, I did the following:
Considered an injective function $f\colon \{s_1,...,s_{n - 1}\} \to \{s_1,...,s_n\}$ given by $s_i \mapsto s_i$,
Induced a group homomorphism between free groups $\phi\colon F(\{s_1,...,s_{n-1}\})\to F(\{s_1,...,s_n\})$ such that $s_i \mapsto s_i$.
Let $N$ be the least normal subgroup of $F(\{s_1,...,s_{n-1}\})$ containing $s^2_i, (s_is_{i + 1})^3$ and $s_is_js^{-1}_is^{-1}_j$ for $|i - j| > 1$. Similarly, let $N'$ be the least normal subgroup of $F(\{s_1,...,s_n\})$ containing $s^2_i, (s_is_{i + 1})^3$ and $s_is_js^{-1}_is^{-1}_j$ for $|i - j| > 1$. Then $\Gamma_n = F(\{s_1,...,s_{n-1}\})/N$ and $\Gamma_{n + 1} = F(\{s_1,...,s_n\})/N'$. It can be shows that $\phi(N) \subseteq N'$, so there is a unique group homomorphism $\psi\colon \Gamma_n\to \Gamma_{n + 1}$ such that $\psi(s_i + N) = s_i + N'$.
The trouble I have is with showing that $\psi$ is injective. If $x,y \in F(\{s_1,...,s_{n-1}\})$ and $x + N' = y + N'$, then I don't see how this implies that $x + N = x + N$. $x + N' = y + N'$ implies that $xy^{-1} = g_1z^{\pm}_1g^{-1}_1...g_kz^{\pm}_kg^{-1}_k$ for some $g_i \in F(\{s_1,...,s_n\})$ and $z_i \in \{ s^2_i, (s_is_{i + 1})^3$ and $s_is_js^{-1}_is^{-1}_j$ for $|i - j| > 1 \mid i = 1,...,n - 1\}$. Of course, $g_1z^{\pm}_1g^{-1}_1...g_kz^{\pm}_kg^{-1}_k$ belongs to $F(\{s_1,...,s_{n -1}\})$ as $xy^{-1}$ does. But how to show that it is an element of $N$?
Also, this seems like there should be a general theorem for dealing with this type of situations. What can it be?
It's clear that $\phi(N) = N' \cap F(\{s_1,\ldots,s_{n-1}\})$ since both are the subgroup $N^*$ of $F(\{s_1,\ldots,s_{n-1}\})$ generated by $\{s_1^2,\ldots,s_{n-1}^2,(s_1s_2)^3,\ldots,(s_{n-1}s_{n-2})^3,[s_1,s_3],\ldots,[s_1,s_{n-1}],[s_2,s_4],\ldots,[s_{n-3},s_{n-1}]\}$, so you're done, as $xy^{-1}$ is clearly in both $N'$ (since $x + N' = y + N'$) and $F(\{s_1,\ldots,s_{n-1}\})$ (since $x,y\in F(\{s_1,\ldots,s_{n-1}\})$.
EDIT: Clarifying why $\phi(N) = N^*$ and $N' \cap F(\{s_1,\ldots,s_{n-1}\}) = N^*$
For $\phi(N)$, note that the generating set that I wrote down is precisely the image of the generating set of $N$ under $\phi$, so $\phi(N)$ and $N^*$ are generated by the same set of generators, so we're done. For $N' \cap F(\{s_1,\ldots,s_{n-1}\})$, note that all of the generators that I've written down lie in $N' \cap F(\{s_1,\ldots,s_{n-1}\})$, which gives us one inclusion straight away. For the other inclusion, note that the only generators we've missed are $s_n^2$, $(s_{n}s_{n-1})^3$, and $[s_i,s_{n}]$ for $i < n - 1$, so we'll add those back in one at a time and show that we don't gain anything in $F(\{s_1,\ldots,s_{n-1}\})$. For $s_n^2$, this is clear, so we have $$\langle N^* \cup \{s_n^2\}\rangle \cap F(\{s_1,\ldots,s_{n-1}\}) = N^*.$$
The others are all similar, so I'll just do $(s_ns_{n-1})^3$. First, note that any element of $$\langle N^* \cup \{(s_ns_{n-1})^3\}\rangle \cap F(\{s_1,\ldots,s_{n-1}\})$$ must contain (as a word with terms in $N^* \cup \{(s_ns_{n-1})^3)$) zero net copies of $(s_ns_{n-1}^3)$ (the $s_n$s have to cancel). In particular, we can pair off each copy $a$ of $(s_ns_{n-1})^3$ with a copy $i(a)$ of $(s_ns_{n-1})^{-3}$ with which it can cancel in some simplification of our word (if we couldn't, we wouldn't be in $F(\{s_1,\ldots,s_{n-1}\})$. Now, everything in between $a$ and $i(a)$ must cancel, so we end up with just the things that aren't in between any $a$ and $i(a)$ - that is, we end up with a word with terms in $N^*$, so $$\langle N^* \cup \{s_n^2\}\rangle \cap F(\{s_1,\ldots,s_{n-1}\}) = N^*.$$ An identical proof will work for all of the $[s_i,s_n]$.