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

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.

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.