Let $X$ be a topological space and $A,B\subseteq X$ subspaces such that $A\subseteq B$ is closed. I've been asked to show that $B/A$ embedds into $X/A$.
The map $\bar{i}:B/A\to X/A$ induced by the inclusion is clearly injective and continuous. To show that it is an embedding I've thought of showing that it is closed.
To do this, let $F\subseteq B/A$ be a closed subspace. I need to show that $\bar{i}(F)$ is closed in $\bar{i}(B/A)$, which is equivalent to show that $\bar{i}(F)=\bar{i}(B/A)\cap \bar{C}$ for some closed subspace $\bar{C}$ of $X/A$.
Let $\pi_B:B\to B/A$ and $\pi_X:X\to X/A$ be the quotient maps. Since $F$ is closed in $B/A$, $\pi_B^{-1}(F)$ is closed in $B/A$, i.e. $\pi_B^{-1}(F)=B\cap C$ for some closed subspace $C$ of $X$. I reckon that $\bar{C}=\pi_X(C)$, but quotient maps are not closed in general.
Is it true that if $A$ is closed then the quotient map is closed?
If not, how can I finish the exercise?
EDIT:
To show that $\bar{C}=\pi_X(C)$ is closed it is enought to show that $\pi_X^{-1}(\bar{C})$ is closed in $X$. But if $C\cap A=\emptyset$, then $\pi_X^{-1}(\bar{C})=C$ and if $C\cap A\neq\emptyset$ then $\pi_X^{-1}(\bar{C})=C\cup A$. In both cases we get closed sets. Is this argument correct?
Proof. Let $F\subseteq X$ be a closed subset. Then $\pi(F)$ is closed in $X/A$ if and only if $\pi^{-1}(\pi(F))$ is closed in $X$. There are two cases
which are both closed. $\Box$
Also I'm not sure what exactly do you mean by "$B/A$ embedds into $X/A$"? $B/A$ is simply a subset of $X/A$. The induced topology agrees with the subspace topology (unless this is what you are after?) and so of course it embedds into $X/A$. It doesn't matter whether $B/A$ is closed in $X/A$ or not, unless you use some custom "embedding" definition.