Any embedding $\mathbb{R} P^n\hookrightarrow \mathbb{R} P^{\infty}$ induces a surjection of the singular cohomology rings.

87 Views Asked by At

On page 79 of "Vector Bundles and K-Theory" Hatcher makes use of the claim in the title where he considers cohomology with integer or $\mathbb{F}_2$ coefficients.

Why is this claim true?

I know that it suffices to see that it induces a surjection on the first cohomology module.

Thanks in advance.

1

There are 1 best solutions below

0
On BEST ANSWER

This is not true for arbitrary embeddings. It is true for embeddings that come from a linear map $\mathbb{R}^n\to\mathbb{R}^\infty$. First, if $i:\mathbb{R}^n\to\mathbb{R}^\infty$ is the standard inclusion then the induced inclusion $\mathbb{R}P^n\to\mathbb{R}P^\infty$ is just the inclusion of the $n$-skeleton for the usual CW-complex structure on $\mathbb{R}P^\infty$ and so the conclusion follows easily using cellular cohomology. Now if $j:\mathbb{R}^n\to\mathbb{R}^\infty$ is any other linear injection, there is a linear automorphism $T:\mathbb{R}^\infty\to\mathbb{R}^\infty$ such that $i=Tj$. Since $T$ induces an homeomorphism from $\mathbb{R}P^\infty$ to itself, the embedding $\mathbb{R}P^n\to\mathbb{R}P^\infty$ induced by $j$ differs from the one induced by $i$ by composition with a homeomorphism, and so also induces surjections on mod $2$ cohomology.

(Alternatively, as Hatcher mentions, you can show that $i$ and $j$ are homotopic through linear injections, which induces a homotopy between the inclusions $\mathbb{R}P^n\to\mathbb{R}P^\infty$. Namely, if $k:\mathbb{R}^n\to\mathbb{R}^\infty$ is a linear injection whose image has trivial intersection with the images of $i$ and $j$, then you can take the straight-line homotopies from $k$ to each of $i$ and $j$.)