Definition on homotopy of maps of pairs and deformation retract of pairs

Question

Reading hatcher's algebraic topology, there is a statement(page127):

Proposition 2.19. If two maps $f, g: (X, A) \rightarrow (Y, B)$ are homotopic through maps of pairs $(X, A) \rightarrow (Y, B)$, then $f_* = g_* : H_n(X, A) \rightarrow H_n(Y, B)$.

Hatcher doesn't give the definition of homotopy between maps of pairs, after some search I guess it should be:

$f, g: (X, A) \rightarrow (Y, B)$ is homotopic if there is some map $H: X \times I \rightarrow Y$ such that:

1. $H(-, 0) = f$
2. $H(-, 1) = g$
3. $H(a, t) \in B$ for all $a \in A$ and $t \in I$.

So question 1: is this definition correct?

question 2 is that I wanna to use proposition 2.19 to prove there is some "natural" map $$ H_n(D, D - \{0\}) \rightarrow H_n(D, \partial D) $$ is a isomorphism for some $n$ dimensional closed ball $D=\{\mathbf{x} \in \Bbb{R}^n|\|x\|_2 \leq 1\}$. So I tried to find some "deformation retract" on the pairs. I tried to mimic some definitions like this:

retract between pairs $r: (X, A) \rightarrow (Y, B)$ is a retract for $Y \subset X$ and $B \subset A$, if $r(y) = y$ for any $y \in Y$ and $r(b) = b$ for any $b \in B$.

(strong) deformation retract between pairs $r: (X, A) \rightarrow (Y, B)$ is a (strong, In other books than Hatcher's) deformation retract if there is some map $F: X \times I \rightarrow X$ such that:

1. $F(-, 0)$ is the identity map of $X$
2. $F(-, 1) = i \circ r$ where $i: (Y, B) \rightarrow (X, A)$ is the inclusion map.
3. $F(y, t) = y$ for all $y \in Y$ and $F(b, t) = b$ for all $b \in B$.

But then it immediately seems that there is no such "deformation retract" between $(D, D - \{0\})$ and $(D, \partial D)$.

So here is question 2: What is the "homotopic equivalent" map between $(D, D - \{0\})$ and $(D, \partial D)$? I think it should intutively related to the deformation retract $D - \{0\}$ and $\partial D$.

And in fact all these questions are from that I wanna to check the following diagram is commutative (where $M$ is a $n-$dimensional manifold and $x, y \in B \subset U$, $B$ and $U$ are homeomorphic to some closed ball. The post here is related to the bottom horizontal isomorphisms. And the vertical arrows are induced by excision theorem I think I have understood. Altough I don't understand why it's commutative yet but I think it should related to naturality and it's another question maybe.)

My original question is that why the bottom horizontal isomorphisms holds. Any idea on this can just ignore the upper questions. Question 1 and 2 is coming from that I tried to prove it with the homotopic equivalent thourgh pairs. To avoid asking an xy question I state my original question here and all these are related to orientation on manifolds (I asked a question about it here relied on the commutative diagram).

Thanks!

Answer 1

1. Your definition of a homotopy of pairs is correct. Hatcher uses the phrase "homotopic through maps of pairs" which means that there is a homotopy $H$ from $f$ to $g$ such that all $H(-,t) : X \to Y$ are maps of pairs, i.e. $H(a,t) \in B$ for all $a \in A$.

2. Indeed there is no reasonable map $(D, D - \{0\}) \to (D, \partial D)$. In particular, there is no "retraction of pairs", let alone a (strong) deformation retraction between pairs. However, you can take the inclusion $j : (D, \partial D) \to (D, D - \{0\})$. This maps induces an isomorphism $$j_*:H_n(D,\partial D)\to H_n(D,D-\{0\})$$ although $j$ is not a homotopy equivalence of pairs (see the related Is $(I^n, \partial I^n)$ homotopy equivalent to $(R^n, R^n\setminus \left\{0\right\})$). To verify this, we will need the Five Lemma (see Hatcher p. 129). In fact we can prove the more general

Theorem: Let $f : (X,A) \to (Y,B)$ be a map of pairs such that both $f : X \to Y$ and $\bar f = f \mid_A :A \to B$ are homotopy equivelences. Then $f_* : H_n(X,A) \to H_n(Y,B)$ is an isomorphism.

Proof. Consider the following commutative diagram:

$\require{AMScd}$ \begin{CD} H_n(A) @>>> H_n(X) @>>> H_n(X,A) @>{\partial}>> H_{n-1}(A) @>>> H_{n-1}(X) \\ @V{\bar f_*}VV @V{f_*}VV @V{f_*}VV @V{\bar f_*}VV @V{f_*}VV \\ H_n(B) @>>> H_n(Y) @>>> H_n(Y,B) @>{\partial}>> H_{n-1}(B) @>>> H_{n-1}(Y) \end{CD} Both rows are exact sequences and all vertical arrows except the middle one are known to be isomorphisms. Now the five lemma shows that also the middle one is an isomorphism.