When is the restriction of a quotient map $p : X \to Y$ to a retract of $X$ again a quotient map?

Question

Quite a number of questions in this forum deal with the problem when the restriction $p' : A \stackrel{p}{\to} p(A)$ of a quotient map $p : X \to Y$ to a subspace $A \subset X$ is again a quotient map. An example is The restriction of a quotient map to a saturated closed set is a quotient map.

Here the question is

Given a quotient map $p : X \to Y$ and a subspace $A \subset X$. When is the restriction $p \mid_A : A \to Y$ a quotient map?

This question occurs in hidden form in the context of projective spaces. Let $\mathbb K = \mathbb R, \mathbb C$. Then $$\mathbb{KP}^n = (\mathbb K^{n+1} \setminus \{0\})/ \sim$$ where the equivalence relation $\sim$ is defined by $x' \sim x$ if $x' = \lambda x$ for some $\lambda \in \mathbb K \setminus \{0\}$. It is well-known that an alternative representation of $\mathbb{KP}^n$ is $$\mathbb{KP}^n = S(\mathbb K^{n+1}) / \sim$$ where $S(\mathbb K^{n+1})$ denotes the unit sphere in $\mathbb K^{n+1}$ (i.e $S(\mathbb R^{n+1}) = S^n$, $S(\mathbb C^{n+1}) = S^{2n+1}$). This shows in particular that $\mathbb{PK}^n$ is compact. Note that for $\mathbb K = \mathbb R$ the relation $\sim$ restricts to $S^n$ via $x' \sim x$ if $x' = \pm x$ and for $\mathbb K = \mathbb C$ restricts to $S^{2n+1}$ via $x' \sim x$ if $x' = \lambda x$ for some $\lambda \in S^1 \subset \mathbb C$.

The above alternative representation of $\mathbb{KP}^n$ is an obvious consequence of the following fact:

Given a map $p : X \to Y$ and a retraction $r : X \to A$ onto a subspace $A \subset X$ such that $p \mid_A \circ \phantom{.} r = p$. Then $p$ is a quotient map if and only if $p \mid_A : A \to Y$ is a quotient map.

The question is to give a proof of this fact to obtain a standard reference in this forum.

Note that $S(\mathbb K^{n+1})$ is a retract of $\mathbb K^{n+1} \setminus \{0\}$; simply define $r : \mathbb K^{n+1} \setminus \{0\} \to S(\mathbb K^{n+1}), r(x) = \dfrac{x}{\lVert x \rVert}$ and note that the quotient map $\pi : \mathbb K^{n+1} \setminus \{0\} \to \mathbb{KP}^n$ has the property $\pi \mid_{S(\mathbb K^{n+1})} \circ \phantom{.} r = \pi$.

Answer 1

Let $p : X \to Y$ be map and $p = q \circ r$ be any factorization with maps $r : X \to Z, q : Z \to Y$.

1. If $p$ is a quotient map, then $q$ is a quotient map.
   Proof. $q$ is surjective because $Y = p(X) = q(r(X)) \subset q(Z)$. Let $U \subset Y$ such that $q^{-1}(U)$ is open in $Z$. Then $p^{-1}(U) = (q \circ r)^{-1}(U) = r^{-1}(q^{-1}(U))$ is open in $X$, thus $U$ is open in $Y$ because $p$ is a quotient map.

2. Assume that $r$ has a right inverse map $i : Z \to X$ (i.e. $r \circ i = id_Z$). Note that this is slightly more general than $r$ being a retraction onto a subspace $Z \subset X$. Then $r$ is a quotient map.
   Proof. $r$ is surjective because $Z = (r \circ i)(Z) = r(i(Z)) \subset r(X)$. Let $V \subset Z$ be such that $r^{-1}(V)$ is open in $X$. Then $V = (r \circ i)^{-1}(V) = i^{-1}(r^{-1}(V))$ is open in $Z$.

3. If $q$ is a quotient map and $r$ has a right inverse map $i : Z \to X$, then $p$ is a quotient map.
   Proof. By 2. $r$ is a quotient map and composition of quotient maps yields a quotient map.