$\mathbb{R}\text{P}^{n-1}$ is not retract of $\mathbb{R}\text{P}^n$

I have to solve the following:

Show that $\mathbb{R}\text{P}^{n-1}$ is not retract of $\mathbb{R}\text{P}^n$ for $n\geq 2$.

I have done this with knowledge of homotopy-groups, by showing that $\mathbb{Z}$ cannot factor through $\mathbb{0}$ or $\mathbb{Z}_{2}$. Yet, I would like to know is there some other way to prove that (without using groups of homotopy)?

Any help is welcome. Thanks in advance.

Solution 1:

One can use the fact that $H^*(\mathbb R P^n, \mathbb Z/2) \cong \mathbb Z/2[x]/x^{n+1}$ as a graded commutative ring, where $x$ is in degree one. The inclusion $\mathbb R P^{n-1} \to \mathbb RP^n$ induces a map of graded rings $\mathbb Z/2[x]/x^{n+1} \to \mathbb Z/2[x]/x^n$. By considering fundamental groups or using the cell structure one can see easily that $x\mapsto x$, and so the map is the standard quotient map. If there were a retraction $\mathbb R P^{n} \to \mathbb RP^{n-1}$, then that would induce a section $\mathbb Z/2[x]/x^{n} \to \mathbb Z/2[x]/x^{n+1}$ of the quotient map. But, then we would still have $x\mapsto x$, and so $0 = x^n \mapsto x^n \neq 0$, a contradiction.