How does one conclude for $S^n\to RP^n$ covering, $0\to H_n(P^n)\to H_n(S^n)\to H_n(P^n)\to\dots$'s morphisms? This is related to Hatcher Algebraic Topology Prop 2B.6. Let $p:S^n\to RP^n$. The coefficient of homology is $Z_2$.
Prop 2B.6 $f:S^n\to S^n$ is an odd map with $f(-x)=-f(x)$ must have odd degree. 
Consider $S^n\to RP^n$ 2 fold covering. Define $\tau: C_\star(P^n)\to C_\star(S^n)$ by lifting simplex $a,b$ s.t. $p(a)=p(b)=x\in C_\star(P^n)$ then $\tau(x)=a+b$. Then clearly $0\to C_\star(P^n)\xrightarrow{\tau}C_\star(S^n)\xrightarrow{p_\star}C_\star(P^n)\to 0$ is exact. 
Clearly this induces long exact sequence $0\to H_n(P^n)\xrightarrow{\tau}H_n(S^n)\xrightarrow{p_\star=0}H_n(P^n)\dots H_1(P^n)\xrightarrow{\partial} H_0(P^n)\xrightarrow{\tau=0}H_0(S^n)\xrightarrow{p_\star}H_0(P^n)\to 0$
$\textbf{Q:}$ Why it is obvious that $p_\star: H_n(S^n)\to H_n(P^n)$ and $\tau: H_0(P^n)\to H_0(S^n)$ are $0$ maps? Note that over $Z_2$, via cellular homology $H_i(P^n)=Z_2$ for $0\leq i\leq n$. My reasoning is to use simplex over $S^n$ which generates $H_n(S^n)$(i.e. Identity map of $S^n\to S^n$ which will vanish in $H_n(P^n)$ by $Z_2$ after $p_\star$ pushing forward.) Similarly $\tau: H_0(P^n)\to H_0(S^n)$ sends the generator to the generator and it is twice. Hence $\tau=0$ at 0-th homology as well. Was there other ways to see this without going through this computation?
 A: From the universal coefficient theorem, if $n$ is odd then the induced map $p_* :S^n \rightarrow P^n$ comes exactly from the map $H^n(S^n) \otimes \mathbb{Z}_2 \rightarrow H_n(P^n) \otimes\mathbb{Z}_2$. It is the case that on homology with coefficients in $\mathbb{Z}$, the map $p_*$ is multiplication by 2. If you are geometrically minded you can look at cellular homology, but you can also work out that the induced map on cohomology is multiplication by 2 through Eilenberg-MacLane spaces and then use the universal coefficient theorem to conclude its map on homology is also multiplication by 2. Either way, the subgroup $2\mathbb{Z} \otimes \mathbb{Z}_2$ is trivial, so the odd case is done.
The even case is similar. The universal coefficient theorem tells you the image of $p_*$ lies in the kernel of the map $H_n(P^n) \rightarrow \operatorname{Tor}(H_{n-1}(P_n),\mathbb{Z}_2)$ which it also says is an isomorphism, so the map is 0. In fact, this argument works for any map $S^n \rightarrow X$ and any coefficient group where $X$ has trivial nth homology.
The map in degree 0 is 0 since the characteristic is 2 and you are lifting an element to 2 times another element since $S^n$ is path-connected.
