# What does a linear map induce on cohomology rings of associated projective spaces?

Suppose $f:\mathbb{R}^n\setminus\{0\}\to \mathbb{R}^m\setminus\{0\}$ is a linear map inducing a map $g:\mathbb{R}P^{n-1}\to \mathbb{R}P^{m-1}$ and hence a homomorphism $$g^*:\mathbb{Z}/2\mathbb{Z}[x]/(x^m)\to \mathbb{Z}/2\mathbb{Z}[y]/(y^n)$$ on cohomology rings with $\mathbb{Z}/2\mathbb{Z}$-coefficients. Since this is a graded ring map, we have $x\mapsto \lambda y$.

Why does linearity imply that $\lambda=1$? Is there an argument without going into the calculation of the cohomology of the projective space?

• Notice that taking lambda equal to 1 does not necessarily give a ring morphism. – Mariano Suárez-Álvarez Jan 29 '17 at 5:23
• I don't really see why... – Antonio Alfieri Jan 29 '17 at 10:30
• @MarianoSuárez-Álvarez Could you please say a little bit more on your comment? Do you mean if $m<n$? – user398460 Jan 29 '17 at 14:39

First of all notice that $g: \mathbb{R}P^n \to \mathbb{R}P^m$ is well defined provided that $f$ is injective. Consequently $m=n+k$ for some $k \geq 0$.
Since the cohomology ring $H^*(\mathbb{R}P^n)=\mathbb{Z}/2\mathbb{Z}[x]/(x^{n+1})$ is generated by the Poincaré dual of (the homology class of) an hyperplane, we have that $$g^*(x)= g^*PD[\text{hyperplane}]=PD[g^{-1}(\text{generic hyperplane})]=y,$$ where the last identity is because the image of $g$ intersects the generic hyperplane in a $n-1$ dimensional subspace.
Another possibility is to prove that $g$ is isotopic to the inclusion $\mathbb{R}P^n \subset \mathbb{R}P^m$.
• ...and... ${}{}$ – Mariano Suárez-Álvarez Jan 29 '17 at 5:22
• Sorry I changed notation here $n=n-1$ and $m=m-1$ – Antonio Alfieri Jan 29 '17 at 10:34
• Well, I don't know. I think he didn't notice that $m\geq n$. – Antonio Alfieri Jan 31 '17 at 8:34