Clarification of the definition of a deformation retract

Question

I found this definition in Munkres:

If I understand correctly, he also makes several claims. The first one is that $r(x)=H(x,1)$ is a retraction of $X$ onto $A$. The second one is that $H$ is a homotopy between $id_X$ and $j\circ r$.

(The first claim says that $r(x)=H(x,1)$. The second claim implies $j(r(x))=H(x,1)$. The codomain of the former map is $A$, the codomain of the latter map is $X$, and formally speaking, these two maps are different. If we restrict the codomain of the latter map, we will get the former map.)

The question is: what for does Munkres talk about $j\circ r$? Why didn't he just say "and $H$ is a homotopy between the identity map of $X$ and the map $r$" instead of the corresponding part of the last sentence in the text? Why consider $j$?

Answer 1

Let $H:X \times I \rightarrow X$ be a homotopy and denote restriction of $H$ on $X \times \{t\}$ by $H_t$ for $t \in [0,1]$. Each $H_t$ can be considered as continuous maps from $X$ into itself as $X \times \{t\}$ is naturally homeomorphic to $X$. Then, $H$ is a homotopy between $H_0$ and $H_1$.

Now in your case, $H(x,1) \in A$ for all $x \in X$, thus $H_1$ is also a continuous map into $A$. Munkres is denoting this map by $r:X \rightarrow A$. Since $r(a)=H(a,1)=a$ for $a\in A$. The map $r$ is a retraction.

The purpose of writing $H_1=j \circ r$ instead of $r$ is that to make sure there is no confusion in the codomain.

Also, observe that $r \circ j:A \rightarrow A$ is the identity map of $A$ and $j \circ r:X \rightarrow X$ is homotopic to the identity map of $X$. Therefore, the map $r:X \rightarrow A$ is a homotopy equivalence with $j:X \rightarrow A$ as its inverse.

Answer 2

The maps aren't different "formally speaking". They are different, with no other qualifier. Two maps with different codomains cannot be equal. It's as simple as that. You need to compose with the inclusion to get a map with the same codomain, and then you can talk about the two maps being equal.

It sounds like you have perfectly understood Munkres's definition, but you are not happy about him being precise for some reason. Math is about precision. If you want hand waving, that's fine, but don't be surprised when you end up proving wrong theorems.

It's also work insisting that the way a subspace is embedded in the ambient space matters. For example, if I tell you that I have a circle embedded in the 3-sphere, then I haven't told you much, and studying what I may have told you is a whole field of algebraic topology called knot theory. When we speak about pairs of spaces, we have a tendency of just writing $(X,A)$, but Munkres is reminding you that the inclusion from $A$ to $X$ is part of the data, which matters much sometimes.