Does $\mathbb R P^n$ retract to $\mathbb R P^k$?

Question

I was trying to prove that no such retraction exists. For $k$ odd this follows directly from the (co-)homology groups of these spaces.

However, for $k$ even I have no idea how to proceed and if it is even true.

Answer 1

This is true as long as $0<k<n$ (it is trivially false for $k=0$ and $k=n$), and follows immediately from the ring structure of the mod 2 cohomology. We have $H^*(\mathbb{R}P^n,\mathbb{Z}/2)=\mathbb{Z}/2[x]/(x^{n+1})$ with $|x|=1$. If $n>k$, then there is no injective homomorphism of graded rings $\mathbb{Z}/2[x]/(x^{k+1})\to \mathbb{Z}/2[x]/(x^{n+1})$ (since $x$ must map to an element $y$ such that $y^{k+1}=0$, and the only such element in degree $1$ is $y=0$).

(I assume you are interested only in the case $k\leq n$, where $\mathbb{R}P^k$ is a subspace of $\mathbb{R}P^n$ in the canonical way. If $k>n$, then $\mathbb{R}P^k$ does not even embed in $\mathbb{R}P^n$.)

Answer 2

Suppose $0<k<n$, $\pi_i(\mathbb{R}P^n)=1$ if $1<i<n$ and $\pi_n(\mathbb{R}P^n)=\mathbb{Z}$. A retract $r:\mathbb{R}P^n\rightarrow \mathbb{R}P^k$ with $i:\mathbb{R}P^k\rightarrow \mathbb{R}P^n$ implies that $(r\circ i)_k=id:\pi_k(\mathbb{R}P^k)\rightarrow \pi_k(\mathbb{R}P^k)$ is trivial. Contradiction.

Suppose that $k>n$, since $H^{k}(\mathbb{R}P^k,\mathbb{Z}/2)=\mathbb{Z/2})$ and $H^{k}(\mathbb{R}P^n,\mathbb{Z}/2)=0$, this implies that $(r\circ i)_k^*=id:H^{k}(\mathbb{R}P^k,\mathbb{Z}/2)\rightarrow H^{k}(\mathbb{R}P^k,\mathbb{Z}/2)$ is trivial contradiction.