# Induction on proposition 2B.6 of Hatcher's Algebraic Topology

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.

• 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. May 15 at 16:40
• 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? May 15 at 16:43
• Any map between path connected spaces induces an isomorphism on $H_0(-)$. May 15 at 18:13