covering map $S^n \rightarrow P^n$ is not null homotopic Here is the problem: Prove that the covering projection $S^n \rightarrow P^n$ is not null-homotopic. This problem is from Algebraic Topology by Harper and Greenberg. There is a suggestion: The lifting theorem provides a set equivalence $[X, S^n] \rightarrow [X, P^n]$ for all simply connected $X$.
How can I prove this? Thank you for your help!
 A: Suppose that the covering map $\pi$ were nullhomotopic through a homotopy $h$. Consider the diagram
$$\begin{matrix}
S^n & \xrightarrow{\mathrm{id}} & S^n\\
i_0\downarrow & & \downarrow \pi\\
S^n\times [0,1]& \xrightarrow{h} & P^n
\end{matrix}$$
which expresses the fact that the homotopy $h$ has a lift at $t=0$. Standard facts about covering spaces tell us that $h$ has a unique lift $\tilde{h}:S^n\times[0,1]\to S^n$ such that $h=\pi\circ\tilde{h}$. Then $\tilde{h}_1:S^n\to S^n$ maps $S^n$ into the fiber over a point $x$ in $P^n$ since $h_1=\pi\circ\tilde{h}_1$ is constant (equal to $x$). Since $S^n$ is connected and the fiber over $x$ is a two point set $\lbrace v,-v\rbrace$ (for some $v\in S^n$) $\tilde{h}_1$ must be constant.
Thus $\iota=\mathrm{id}_{S^n}$ is homotopic to a constant map $S^n\to\mathrm{pt}\to S^n$ (equal to $\pm v$). This poses a problem when passing to (reduced) homology (for simplicity) since we must have that $\iota_*=\mathrm{id}_{\tilde{H}_*(S^n)}=0$ since it factors through the reduced homology of a point (which is $0$) contradicting the fact that $\tilde{H}_*(S^n)\simeq\Bbb Z[n]$ (an infinite cyclic group concentrated in degree $n$).
A: As you've accepted an answer, it's probably ok to post an answer which uses a sledgehammer style proof.
The covering projection $p\colon S^n\to P^n$ is a fibration with fiber homotopy equivalent to $\mathbb{Z}_2$, and in particular there is a long exact sequence $$\cdots\to\pi_n(\mathbb{Z}_2)\to\pi_n(S^n)\stackrel{p_*}{\to}\pi_n(P^n)\to\pi_{n-1}(\mathbb{Z}_2)\to\cdots$$ associated to this fibration. For $n\geq 1$, $\pi_n(\mathbb{Z}_2)$ is trivial and so $p_*$ is a monomorphism from a non-trivial (infinite cyclic) group into $\pi_n(P^n)$ and so is not the trivial map. It follows that $p$ is not null-homotopic.
As Qiaochu said in the comments, for $k\geq 2$ we also have $\pi_{k-1}(\mathbb{Z}_2)=0$ and so we even know that $p_*\colon\pi_k(S^n)\to\pi_k(P^n)$ is an isomorphism for all $k\geq 2$, $n\geq 1$.
A: A slightly different approach: the homotopy cofiber of $S^n \to \mathbb RP^n$ is $\mathbb RP^{n+1}$. Thus, if $S^n \to \mathbb RP^n$ were nullhomotopic then $RP^{n+1} \simeq \mathbb RP^n \vee S^{n+1}$, but that is not the case by taking, for example, cohomology with $\mathbb Z/2$ coefficents. To wit: $H^*(\mathbb RP^n \vee S^{n+1}, \mathbb Z/2) = \mathbb Z/2[x_1,y_{n+1}]/(x_1^n, x_1y_{n+1}, y_{n+1}^2)$, whereas $H^*(\mathbb RP^{n+1}) = \mathbb Z/2[x_1]/(x_1^{n+1})$
