So proposition 2B.6 states that an odd map $f:S^n\rightarrow S^n$ must have odd degree. I've understood (perhaps incorrectly) that the following diagram of short exact sequences seen here induces a map between the long exact sequence seen here. I'm using $\varphi$ to indicate the map induced in $\mathbb{R}P^n$ by $f$. Hatcher states that we can see that all the maps in the long sequence are isomorphisms by induction on dimension, starting with the fact that $f_*$ and $\varphi_*$ are isomorphisms in dimension zero.

I could see why $\varphi_*$ would be an isomorphism if "dimension zero" referred to $\mathbb{R}P^0$, since that's just a point, but then I can't figure out how the induction step would be done to connect the result from $\mathbb{R}P^{n-1}$ to $\mathbb{R}P^n$, and if "dimension zero" just refers to $H_0(\mathbb{R}P^n,\mathbb{F}_2)$ and the induction refers to just following along the squares in that long exact sequence, then I really don't see why it would be obvious that $\varphi_*:H_0(\mathbb{R}P^n,\mathbb{F}_2)\rightarrow H_0(\mathbb{R}P^n,\mathbb{F}_2)$ is an isomorphism.

  • 1
    $\begingroup$ The "dimension zero" refers to the grading of the long exact sequence. Hatcher instructs you to observe that if three maps in a commutative square are isomorphisms, then so is the fourth. Apply this to the commutative squares resulting from the naturality of the boundary maps. This yields the desired induction. $\endgroup$
    – Thorgott
    May 15 at 16:40
  • $\begingroup$ Thank you, but also, how do I start that induction? I understood that if I prove that the map from $H_0(P^n)$ to itself is an isomorphism than all others maps will be too, but how do I prove that? $\endgroup$ May 15 at 16:43
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    $\begingroup$ Any map between path connected spaces induces an isomorphism on $H_0(-)$. $\endgroup$ May 15 at 18:13


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