I'm working on Hatcher Example 2.23 and I'm having a little trouble understanding the argument.
Let us find explicit cycles representing generators of the infinite cyclic groups $H_n\left(D^n, \partial D^n\right)$ and $\tilde H_n(S^n)$. Replacing $\left(D^n, \partial D^n\right)$ by the equivalent pair $\left(\Delta^n, \partial \Delta^n\right)$, we will show by induction on $n$ that the identity map $i_n: \Delta^n \rightarrow \Delta^n$, viewed as a singular $n$-simplex, is a cycle generating $H_n\left(\Delta^n, \partial \Delta^n\right)$. That it is a cycle is clear since we are considering relative homology. When $n=0$ it certainly represents a generator. For the induction step, let $\Lambda \subset \Delta^n$ be the union of all but one of the $(n-1)$-dimensional faces of $\Delta^n$. Then we claim there are isomorphisms
$$H_n\left(\Delta^n, \partial \Delta^n\right) \stackrel{\approx}{\longrightarrow} H_{n-1}\left(\partial \Delta^n, \Lambda\right) \stackrel{\approx}{\longleftarrow} H_{n-1}\left(\Delta^{n-1}, \partial \Delta^{n-1}\right) $$
The first isomorphism is a boundary map in the long exact sequence of the triple $\left(\Delta^n, \partial \Delta^n, \Lambda\right)$, whose third terms $H_i\left(\Delta^n, \Lambda\right)$ are zero since $\Delta^n$ deformation retracts onto $\Lambda$, hence $\left(\Delta^n, \Lambda\right) \simeq(\Lambda, \Lambda)$. The second isomorphism comes from the preceding proposition since we are dealing with good pairs and the inclusion $\Delta^{n-1} \hookrightarrow \partial \Delta^n$ as the face not contained in $\Lambda$ induces a homeomorphism of quotients $\Delta^{n-1} / \partial \Delta^{n-1} \approx$ $\partial \Delta^n / \Lambda$. The induction step then follows since the cycle $i_n$ is sent under the first isomorphism to the cycle $\partial i_n$ which equals $\pm i_{n-1}$ in $C_{n-1}\left(\partial \Delta^n, \Lambda\right)$.
The first part of the argument makes sense, but I'm a little confused by how the induction step follows: when is the second isomorphism used? How do I know where the cycle $\pm i_{n-1}$ is sent to in the third group?
I guess my general question is: suppose $f:(X,A)\to (Y,B)$ induces a homeomorphism on quoteints: $X/A\approx Y/B$, then it induces an isomorphism between relative homology groups:
$$
H_n(X,A)=\widetilde H_n(X/A)=\widetilde H_n(Y/B)=H_n(Y,B).
$$
But how do I understand this map $H_n(X,A)\to H_n(Y,B)$? For instance, what does it do to the generator?
A similar argument (I think) is used in Theorem 2.27, proving the equivalence of singular and simplicial homology.
Any help is appreciated! Thank you!
Best Answer
Your general question
Suppose that $(X,A)$ and $(Y,B)$ are good pairs, and suppose that $f: (X, A) \to (Y, B)$ is a map of pairs, which induces a homeomorphism $\overline{f} : X/A \overset{\cong}{\to} Y / B$.
Let $p_X: X \to X/A$ and $p_Y : Y \to Y/B$ be the natural quotient maps, and let $\text{id}_{X/A} : X/A \to X/A$ and $\text{id}_{X/B} : X/B \to X/B$ be the identity maps on $X/A$ and $Y/B$ respectively.
The key insight is that we have a commutative diagram of maps of pairs. $\require{AMScd}$ \begin{CD} (X, A) @>{f}>> (Y, B)\\ @V{p_X}VV @V{p_Y}V{}V \\ (X/A, A/A) @>{\overline{f}}>{}> (Y/B, B/B) \\ @A{\text{id}_{X/A}}A{}A @A{\text{id}_{Y/B}}A{}A \\ (X/A, \emptyset) @>{\overline{f}}>{\cong}> (Y/B, \emptyset) \end{CD}
These maps of pairs induce homomorphisms between homology groups, which also form a commutative diagram. $\require{AMScd}$ \begin{CD} H_n(X, A) @>{f_\star}>> H_n(Y, B)\\ @V{(p_X)_\star}V{\cong}V @V{(p_Y)_\star}V{\cong}V \\ H_n (X/A, A/A) @>{\overline{f}_\star}>{}> H_n (Y/B, B/B) \\ @A{(\text{id}_{X/A})_\star}A{\cong}A @A{(\text{id}_{Y/B})_\star}A{\cong}A \\ \widetilde{H}_n (X/A) @>{\overline{f}_\star}>{\cong}> \widetilde H_n (Y/B) \end{CD}
To spell this out:
Using the bottom square in the commutative diagram, we can deduce that the homomorphism $$\overline{f}_\star: H_n(X/A, A/A) \to H_n(Y/B, B/B)$$ is an isomorphism.
The top square in the commutative diagram then tells us that the homomorphism $$f_\star: H_n(X,A) \to H_n(Y,B)$$ is an isomorphism.
Now, let's use this information to answer your question.
By our reasoning above, this isomorphism $H_n(X,A)\to H_n(Y,B)$ is precisely the homomorphism $f_\star: H_n(X,A) \to H_n(Y,B)$!
Therefore, if $[\sigma]$ is a generator of $H_n(X,A)$, then the image of this generator under the isomorphism $H_n(X,A)\to H_n(Y,B)$ is the element $f_\star[\sigma] = [f \circ \sigma]$.
Your specific question about Hatcher's Example 2.23
The "second isomorphism" is the homomorphism $$ \iota_\star: H_{n-1}\left(\Delta^{n-1}, \partial \Delta^{n-1}\right) \to H_{n-1}\left(\partial \Delta^n, \Lambda\right),$$ where $\iota: \Delta^{n-1} \to \partial \Delta^n$ is the inclusion map sending $\Delta^{n-1}$ to the face of $\partial \Delta^n$ that is not contained in $\Lambda$.
You ask the question:
Based on our discussion above, the answer is that $i_{n-1}$ is sent to $\iota_\star (i_{n-1})$!
To be completely explicit: