induced maps in homology

Question

Let $S^1=I^1/ \partial I^1$, where $I^1=[0,1]$ with base point ${0}$ and $I^n=I^1\times I^1 \times \dots \times I^1$ (n times). $S^p=S^1 \wedge \dots \wedge S^1 $ where $\wedge$ is the smash product. Now we define $\gamma_1:I^1 \to S^1$ to be the quotient map. And $\gamma_p=\gamma_1 \wedge \dots \wedge \gamma_1$ (p times).

Consider the map $1 \wedge \gamma_p:I^{p+1} \to I^1 \wedge S^p$. Now we restrict this map to the boundary $\partial I^{p+1}$. Now we want to shot that $(1 \wedge \gamma_p)(\partial I^{p+1})=S^p$ In the book I'm reading (Bredon's topology and geometry) I found the following calculation I don't understand. I put a ? at the place I don't understand. $\partial I^{p+1}=(\partial I^1)\times I^p \bigcup I^1 \times(\partial I^p) \to (\partial I^1)\times S^p \bigcup I \times \{ {\star} \} \to^? S^0 \wedge S^p=S^p$

Then Bredon also claims that $1 \wedge \gamma_p:\partial I^{p+1} \to S^p $ induces the isomorphism in homology: $H_p(\partial I^{p+1},\star) \to H_p(S^p,\star)$ Why? From the lecture I know that if f is a homotopy equivalence it induces an isomorphism in homology. So I thought that probably $(1 \wedge \gamma_p)$ is a homotopy equivalence?

And my last question: I have a map (induced by the connecting homomorphism) $\partial_{\star}:H_{p+1}(I^{p+1},\partial I^{p+1}) \to H_p(\partial I^{p+1},\star)$. Why is this map an isomorphism? By the axioms it is only a homomorphism.

Answer 1

OP asked three questions, I think I'm able to answer the 1st and the 3rd, but also confused with the 2nd, so I'd appreciate it if someone can complete my following answer.

For the first question, the last arrow you are asking is the quotient map collapsing $I^1\vee S^p$ to a point. To see this we can first write down the construction of $1 \wedge \gamma_p:I^{p+1} \to I^1 \wedge S^p$ step by step and see how the restriction of domain into $\partial I^{p+1}$ changes each step. We have $$1 \wedge \gamma_p:I^{p+1} =I^1\times I^p \to I^1 \times S^p\to (I^1 \times S^p)/(I^1\vee S^p)=I^1\wedge S^p,$$ where the first arrow is the map $1\times \gamma_p$, the second is natually the quotient map collapsing $I^1\vee S^p$ to a point. Now you can easily trace how $1 \wedge \gamma_p|_{\partial I^{p+1}}$ behaves at each step, and it turns out it is still the same collapsing map. To see a concrete example, consider doing the above to $1 \wedge \gamma_1$:

where the upper is the construction of $1 \wedge \gamma_1$ and lower is the restriction $1 \wedge \gamma_1|_{\partial I^2}$. Everything should be intuitively clear.

For the third question, just look at the proof of Theorem 6.6 in the same chapter of Bredon, the part proving $(D_n)\iff(S_{n-1})$, it's the same thing except here $S^p$ is written as $\partial I^{p+1}$, $D^p$ as $I^p$, and together with the fact that $H_n(X,*)=\tilde{H}_n(X)$.

As a comment to the OP's second question, I would like to mention that Bredon, after briefly describing what the map $1 \wedge \gamma_p|_{\partial I^p}$ does geometrically, explicitly stated that

$\ldots$Clearly it is a homotopy equivalence.(Indeed, it is homotopic to a homeomorphism.)

But I just can't see how it works.