Use homotopy groups to show there is no retraction \(\mathbb{RP}^n \to \mathbb{RP}^k\) if \(n > k > 0\).
Solution
Suppose there were a retraction \[ r:\mathbb{RP}^n\to\mathbb{RP}^k, \qquad n>k>0, \] of the standard inclusion \(i:\mathbb{RP}^k\hookrightarrow\mathbb{RP}^n\). Then \[ r_*i_*=\operatorname{id} \] on every homotopy group, so \(i_*\) would be injective.
If \(k\ge2\), covering-space invariance of higher homotopy groups gives \[ \pi_k(\mathbb{RP}^k)\cong\pi_k(S^k)\cong\mathbb Z, \] whereas, since \(k<n\), \[ \pi_k(\mathbb{RP}^n)\cong\pi_k(S^n)=0. \] Thus \(i_*:\mathbb Z\to0\) cannot be injective.
If \(k=1\), then \(\mathbb{RP}^1\cong S^1\), so \[ \pi_1(\mathbb{RP}^1)\cong\mathbb Z, \qquad \pi_1(\mathbb{RP}^n)\cong\mathbb Z_2 \quad(n>1). \] Again no injection \(\mathbb Z\hookrightarrow\mathbb Z_2\) exists.
Therefore \[ \boxed{\mathbb{RP}^n\text{ does not retract onto }\mathbb{RP}^k\text{ for }n>k>0.} \]