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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.

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    $\begingroup$ I don’t think this is true. Finite dimensional projective space can be embedded in high enough Euclidean space. Since infinite projective space contains copies of arbitrarily large Euclidean space we can embed each projective space null homotopically. Perhaps there is a restriction on cellularity of the embedding. $\endgroup$ Aug 10, 2019 at 21:19

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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 $j$.)

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