Question on Proposition 2.22 of Hatcher’s Algebraic Topology: How to induce a deformation retraction on the quotient space

algebraic-topologygeneral-topology

Let $X$ be a topological space and $A$ is a subspace of $X$.
Proposition 2.22 in Hatcher's Algebraic Topology describes the relation of relative of the homology $H_n(X/A)$ of the quotient space $X/A$ and the singular homology $H_n(X, A)$.

Proposition 2.22. For good pairs $(X,A)$, the quotient map $q: (X,A) \rightarrow (X/A, A/A)$ induces isomorphisms
$$q_*: H_n (X,A) \rightarrow H_n(X/A, A/A)$$
for all $n$.

Here, “good pairs'' means that $A$ is a nonempty closed subspace and is a deformation
retract of some neighborhood in $X$.

Let $V$ be a neighbourhood of $A$ satisfying the condition above.
In the proof of Proposition 2.22, the author claims that the deformation retraction of $V$ onto $A$ induces a deformation retraction of $V/A$ onto $A/A$.

I think I can figure out what the induced deformation retraction of $V/A$ onto $A/A$ is, but I do not know why it is indeed a deformation retraction, because I do not know how to prove the corresponding homotopy map is continuous.

For details, let $i: A \rightarrow V$ be the inclusion map.
Since $A$ is a deformation retract of $V$, there is a continuous map $r: V \rightarrow A$ and a continuous map $F(x,t): V \times I \rightarrow V$ such that
$$r(a) =a, \quad F(x,0) =\operatorname{id}_V (x) =x, \quad F(x,1) =r(x), \quad F(a,t) =a, \quad \forall x \in V,\ \forall t \in I,\ \forall a \in A,$$
where $I=[0,1]$.

Let $q: V \rightarrow V/A$ be the quotient map, i.e.,
$$q(x) =\begin{cases}
A & x \in A \\ \{ x \} & x \not\in A
\end{cases}$$
Let $\tilde{i}: A/A \rightarrow V/A$ be the inclusion map.
The induced deformation retraction $\tilde{r}: V/A \rightarrow A/A$ must be defined by $q(x) \mapsto A$ since $A/A$ consists only one element $A$.
But why the composition map $\tilde{i} \circ \tilde{r}$ is homotopic to $\operatorname{id}_{V/A}$ relative to $A/A$?

A natural construction of the homotopy map from $\operatorname{id}_{V/A}$ to $\tilde{i} \circ \tilde{r}$ seems to be given by letting $\tilde{F}: V/A \times I \rightarrow V/A$ be
$$\tilde{F} (q(x), t) =q(F(x,t)).$$

I can check all the conditions we need except the one that $\tilde{F}$ is continuous.

We can choose a closed subset $Z$ in $V/A$.
Then we see that
$$\tilde{F}^{-1} (Z) =(q \times \operatorname{id}_{I}) (F^{-1} (q^{-1} (Z))),$$
where $q \times \operatorname{id}_I: V \times I \rightarrow V/A \times I$ is the product of the quotient $q: V \rightarrow V/A$ and the identity map of $I$.
Since $F$ and $q$ are both continuous, we obtain a closed subset $F^{-1} (q^{-1} (Z))$ in $V \times I$.

Although $q$ is a closed map because $A$ is a closed subset, the product map $q \times \operatorname{id}_I$ is not necessarily a closed map, so $\tilde{F}^{-1} (Z)$ is not necessarily a closed subset (I think).
Where am I going wrong?

Thank you very much for reading this rambling (and maybe trivial) question.

Best Answer

All you have to know is that if $p : Y \to Z$ is a quotient map, then also $p \times id_I : Y \times I \to Z \times I$ is a quotient map. This is a well-known theorem from general topology.

In particular $\tilde q = q \times id_I : V \times I \to (V/A) \times I$ is a quotient map. Your definition of $\tilde F$ says that $\tilde F \circ \tilde q = q \circ F$. Since $q \circ F$ is continuous, the universal property of quotient maps implies that $\tilde F$ is continuous.

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