The restriction of the quotient map $q:X\rightarrow X/A$ to the subspace $X\setminus A$ is a homeomorphism As part of a proof regarding reduced homology, I need to know that the following fact holds for good pairs $(X,A)$:
If $q:X\rightarrow X/A$ is the quotient map, then  $q|_{X\setminus A}:X\setminus A \rightarrow (X/A)\setminus (A/A)$ is a homeomorphism.
I can see why the restriction is injective, surjective (by the definitions) and continuous (since it is a restriction of a continuous map), but I can't see why the inverse is also continuous.
Question 1: How can I prove that the inverse is continuous?
Question 2: Is this in general true, when $A$ is just a subspace of $X$?
Thank you very much
*** A good pair of spaces $(X,A)$ is a space $X$ along with a subspace $A$, such that there exists an open subset $U$ of $X$ which contains the closure of $A$ such that the inclusion $i:A\rightarrow U$ is a deformation retraction.
Comment: it might be the case that $A$ is closed or open, but the lecturer didn't mention that. I will ask him and edit the post accordingly.
 A: The following appears to be a minimal counterexample.
Let $X = \{ 1,2,3,4 \}$ with the topology $\{ \emptyset, \{ 3,4 \}, X \}$ and let $A = \{ 2,3 \}$. It is clear
that $\{4\}$ is open in $X\setminus A$. On the other hand
$X/A$ has the trivial topology, since $\{3,4\}$ is not a union of equivalence classes. Therefore $(X/A) \setminus (A/A)$ also has the 
trivial topology.
To see that $A$ is a deformation retract of $X$, we can define the 
homotopy $h: X \times [0, 1] \to X$ as follows:
$$
\begin{eqnarray}
h(1,t)&=& \cases{2 & if t = 1, \\
                 1 & otherwise;
           } \\
h(2,t)&=& 2 \\
h(3,t)&=& 3 \\
h(4,t)&=& \cases{4 & if t = 0, \\
                 3 & otherwise.
          }
\end{eqnarray}
$$
This is continuous, since 
$h^{-1}[\{3,4\}] = \{3, 4\} \times [0,1]$. If you would prefer a T0
counterexample, this can be obtained with only a little more work by adding the open sets $\{ \{1\}, \{1,2,3\}, \{3\} \}$.
However if we assume that $X$ is a Hausdorff space, then $A$ is closed
in some neighbourhood U, by virtue of being a retract. Since by hypothesis
$\overline{A} \subset U$, it must also be closed in $X$. In that case a
standard result applies: if $q: X \to Y$ is a quotient map and $F \subset Y$ is closed or open, then $q|_{q^{-1}[F]}^F$ is also a quotient map.
A: Thinking out loud:
So you're identifying $A$ to a point, so define $R_A$ to be the unique equivalence relation with classes $\{\{x\}: x \notin A\} \cup \{A\}$ or $xR_A y $ iff ($x \in A \land y \in A$) or ($x=y$). So $X/A$ (the set of aforementioned classes) gets the quotient topology w.r.t. $q; X \rightarrow X/A$ that sends each $x \notin A$ to $\{x\}$ and $x \in A$ to $A$. Now consider $q' = q | X\setminus A$ which maps $X \setminus A$ to $(X/A) \setminus \{\{A\}\}$, using $q(x) = \{x\}$ as said.
Then $q'$ is 1-1 on $X \setminus A$, and onto its codomain, and $q'$ is clearly continuous as the restriction of a continuous map. Now if $O$ is open in $X \setminus A$, so $O = U \cap (X\setminus A)$ for some $U$ open in $X$, then we need to show that $q'[O]$ is open in its image, so there must be some open $U' \subset X / A$ such that $U' \cap (X/A \setminus \{\{A\}\}) = U' \setminus \{\{A\}\} = q'[O]$. This $U'$ must satisfy that $q^{-1}[U']$ open in $X$. $q^{-1}[U'] = O \cup A$, if $\{A\} \in U'$ and $O$ otherwise. But these need not be open in general. So I'm not sure your statement is even true. 
For $A$ closed or open it's clear as then $q$ is a closed (resp. open) map, and so hereditarily quotient. For general $A$?
