Can we guarantee that $F/H$ is a free group? Let $F$ be the free group on the set $X$, and let $Y\subseteq X$.
Suppose that $H$ is the smallest normal subgroup of $F$ containing $Y$.
Can we guarantee that $F/H$ is a free group?  
If the answer is affirmative, can someone suggest a proof?
 A: You can prove that $F/H$ has the universal property of free groups with respect to a particular set. Consider $Z=X\setminus Y$, an arbitrary group $G$ and a map $f\colon Z\to G$: we wish to see that $f$ extends uniquely to a group homomorphism $\hat{f}\colon F/H\to G$ such that, for every $z\in Z$,
$$
\hat{f}(zH)=f(z)
$$
Define first $g\colon X\to G$ by
$$
g(x)=\begin{cases}
f(x) & \text{if $x\in Z$}\\[4px]
1 & \text{if $x\in Y$}
\end{cases}
$$
By the universal property of $F$, the map $g$ extends uniquely to a group homomorphism $\hat{g}\colon F\to G$ such that, for every $x\in X$, $\hat{g}(x)=g(x)$.
Now, if $x\in Y$, we have $\hat{g}(y)=g(y)=1$, so $Y\subseteq\ker\hat{g}$ and therefore $H\subseteq\ker\hat{g}$. Hence $\hat{g}$ induces a unique homomorphism $\hat{f}\colon F/H\to G$ such that, for every $a\in F$,
$$
\hat{f}(aH)=\hat{g}(a)
$$
In particular, if $z\in Z$, $\hat{f}(zH)=\hat{g}(z)=g(z)=f(z)$, as required.
The only thing to check now is that the map $z\mapsto zH$ is injective.
Suppose $z_1H=z_2H$ for $z_1,z_2\in Z$, $z_1\ne z_2$. Consider the map $h\colon X\to\{1,w,w^2\}$ (a cyclic group with three elements) defined by $h(z_1)=w$, $h(z_2)=w^2$ and $h(x)=1$, for all other elements of $X$.
Since $F$ is free on $X$, $h$ extends uniquely to a group homomorphism $\hat{h}\colon F\to\{1,w,w^2\}$, whose kernel contains $Y$ and hence $H$. Now
$$
\hat{h}(z_1)=w\ne w^2=\hat{h}(z_2)
$$
so $z_1z_2^{-1}\notin\ker\hat{h}$ and, in particular, $z_1z_2^{-1}\notin H$.
A: Suppose first $\;X=\{x_1,...,x_n\}\;$ is finite, and for the sake of simple things assume $\;Y=\{x_1,...,x_k\}\;,\;\;1\le k<n\;$ . Define
$$\phi: F_n:=F(X)\to F_{n-k}:=F(x_{k+1},...,x_n)\;,\;\;\phi w(x_1,...,x_n):=w(1,1,...,1,x_{k+1},...,x_n)$$
with $\;w\in F_n\;$ a reduced word . This homomorphism is the unique one extending the function 
$$f:X\to Y\;,\;\;f(x_i)=\begin{cases}1\;,&1\le i\le k\\{}\\x_i,\,&k+1\le i\le n\end{cases}$$
by the universal property of free groups.
Now, if $\;h:=r^{-1}qr\;,\;\;r\in F_n\;,\;\;q\in F_{n-k}\;$ , then 
$$\phi h=\phi(r)^{-1}\cdot1\cdot\phi(h)=1\implies \langle Y\rangle^{F_n}\le\ker\phi$$
and on the other hand, if $\;\phi w=1\;$ , then: if the last letter in the word $\;w\;$ is in $\; Y\;$, then a simple inductive argument on the length of $\;w\;$ gives us that $\;w\in\langle Y\rangle^{F_n}\;$, otherwise: we can write
$$w=w'x_m\;,\;\;k+1\le m\le n\implies 1=\phi w=\phi w'\cdot x_m\implies x_m^{-1}$$
must appear in $\;\phi w'\;$ , so we can in fact write $\;w_1x_m^{-1}w_2=w\;$ , and then
$$w=w_1x_m^{-1}w_2x_m$$
and again an inductive argument shows $\;w\in\langle Y\rangle^{F_n}\;$.
.
I came up alone with the above. Hopefully someone more expert than I can check it.
The above shows $\;F_n/\langle Y\rangle^{F_n}\cong F_{n-k}\;$
The infinite case is very similar but I haven't thought it through completely, though I think the difference is very small.
