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The Problem: Using the cup product structure, show there is no map $\mathbb{R}P^n \rightarrow \mathbb{R}P^m$ inducing a nontrivial map $H^1(\mathbb{R}P^m; \mathbb{Z_2}) \rightarrow H^1(\mathbb{R}P^n; \mathbb{Z_2})$ if $n > m$. What is the corresponding result for maps $\mathbb{C}P^n \rightarrow \mathbb{C}P^n$?

Notes: For cochains $\phi \in C^k(X; R)$ and $\psi \in C^l(X; R)$ we can define the cup product $\phi \smile \psi \in C^{k+l}(X;R)$ is the cochain defined on the singular simplex $\sigma: \Delta^{k+l} \rightarrow X$ is given by the formula: $$\phi \smile \psi(\sigma) = \phi(\sigma|[v_0,...,v_k])\psi(\sigma|[v_k,...,v_{k+l}])$$ There are some properties of these cochairs for instance we have the induced cup product $H^k(X; R) \times H^l(X; R) \xrightarrow{\smile} H^{k+l}(X; R)$. I'm not sure if the induced map in this problem is the same as this one. I suspect this problem has something to do with Theorem 3.12 of this section because that theorem is about the projective space but it only contains $H^*$ which is defined to be the direct sum of all homology groups. I'm having trouble finding in this section information that relates the first cohomology group of $\mathbb{R}P^m$ and $\mathbb{R}P^n$.

Thank you!

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    $\begingroup$ Do you know how to compute cup products on $\mathbb{R}P^n$? $\endgroup$ – Jason DeVito Aug 27 '20 at 20:24
  • $\begingroup$ Correct me if I'm wrong but in this section it only gives a very general formula for this product. I'm not sure how to apply it to $\mathbb{R}P^n$. $\endgroup$ – Math_Day Aug 27 '20 at 20:38
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    $\begingroup$ Use Hatcher's Theorem 3.12. $\endgroup$ – John Palmieri Aug 27 '20 at 20:41
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The only nontrivial map $H^1(\mathbb{R}P^m;\mathbb{Z}/(2))\to H^1(\mathbb{R}P^n;\mathbb{Z}/(2))$ is the identity since both domain and codomain is isomorphic to $\mathbb{Z}/(2)$. Assume for contradiction that $f:\mathbb{R}P^n\to \mathbb{R}P^m$ induces this nontrivial map on $H^1$.

Recall that $H^*(\mathbb{R}P^n;\mathbb{Z}/(2))\cong \mathbb{Z}/(2)[x]/(x^{n+1})$ with $|x|=1$, and that the induced map $f^*:H^*(\mathbb{R}P^m;\mathbb{Z}/(2))\to H^*(\mathbb{R}P^n;\mathbb{Z}/(2))$ is a ring homomorphism. This gives us that we have ring homomorphism $f^*:\mathbb{Z}/(2)[x]/(x^{m+1})\to \mathbb{Z}/(2)[x]/(x^{n+1})$ which maps $x\to x$. However, this is not a ring homomorphism since $n>m$, hence we have arrived at a contradiction. We conclude that no such $f$ exists.

The similar result for $\mathbb{C}P^n$ is that there exists no map $\mathbb{C}P^n\to \mathbb{C}P^m$ inducing a nontrivial map $H^2(\mathbb{C}P^m;\mathbb{Z})\to H^2(\mathbb{C}P^n;\mathbb{Z})$ when $n>m$. The argument for this is the same mutatis mutandis. Note in this case the nontrivial map on $H^2$ does not need to be the identity since we are working with coefficients over $\mathbb{Z}$ now. However, it is still multiplication by a non-zero integer which still leads to the contradiction.

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  • $\begingroup$ How do you know we cannot have a homomorphism between $\mathbb{Z}(2)[x]/(x^{m+1}) \rightarrow \mathbb{Z}(2)[x]/(x^{n+1})$? If $m+1|n+1$ then we could map it like $x \rightarrow x^\frac{n+1}{m+1}$. $\endgroup$ – Math_Day Aug 29 '20 at 0:19
  • $\begingroup$ It is possible to get such a homomorphism, but $f^*$ can never be this morphism since it is forced to be graded. In this case this means that $x\mapsto \lambda_1 x$, $x^2\mapsto \lambda_2 x^2$, $x^3 \mapsto \lambda_3 x^3,\ldots$, where $\lambda_i\in \mathbb{Z}/(2)$. $\endgroup$ – Frederik Aug 29 '20 at 16:47
  • $\begingroup$ Also recall that $x$ in this case is the generator of the cohomology group $H^1(\mathbb{R}P^m;\mathbb{Z}/(2))$ in the domain and the generator of $H^1(\mathbb{R}P^n;\mathbb{Z}/(2))$ in the codomain. That $f$ induces isomorphism on $H^1$ precisely means that $f^*:x\mapsto x$. $\endgroup$ – Frederik Aug 29 '20 at 16:49

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