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We know the cohomology ring with coefficients in $\mathbb{Z}/2$ of projective real space $\mathbb{PR}^n$ is

$$H^*(\mathbb{PR}^n, \mathbb{Z}/2) = \mathbb{Z}/2[X]/(X^{n+1})$$

with graduated generator $X \in H^1(\mathbb{PR}^n, \mathbb{Z}/2)$. Futhermore, by Künneth for cohomology we also have isomomorpism of graduated rings

$$H^*(\mathbb{PR}^n \times \mathbb{PR}^m, \mathbb{Z}/2) \cong H^*(\mathbb{PR}^n, \mathbb{Z}/2) \otimes H^*(\mathbb{PR}^m, \mathbb{Z}/2) \cong \mathbb{Z}/2[a]/(a^{n+1}) \otimes \mathbb{Z}/2[b]/(b^{m+1}) \cong \mathbb{Z}/2[a, b]/(a^{n+1}, b^{n+1})$$

the last one comes with identifications $a \otimes 1 \mapsto a, 1 \otimes b \mapsto b$.

From now we consider $\mathbb{PR}^n \cong (\mathbb{PR}^n, pt)$ as pointed space. Therfore we can define following canonical morphisms:

$i_n:\mathbb{PR}^n \times \{pt\} \hookrightarrow \mathbb{PR}^n \times \mathbb{PR}^m $

$pr_n: \mathbb{PR}^n \times \mathbb{PR}^m \to \mathbb{PR}^n \times \{pt\}$.

Analogously $i_m, pr_m$ for $\{pt\} \times \mathbb{PR}^m$. Let us now consider the induced morphisms $i_n^*$ and $pr_n^*$ on cohomology. Cince $pr_m \circ i_n, pr_n \circ i_m$ are constant maps and $pr_n \circ i_n = id_{\mathbb{PR}^n}$, $pr_m \circ i_m = id_{\mathbb{PR}^m}$ und $H^*(-)$ is functorial the compostions above become zero maps resp identity on corresponding cohomology rings.

My question is the following:

I don't know how to use the given identity $ i_k ^* \circ pr_l ^* = \delta_{k,l} Id $ for $k,l \in \{m,n\}$

to derive explicitely that

$i_n ^*:H^*(\mathbb{PR}^n \times \mathbb{PR}^m, \mathbb{Z}/2) \to H^*(\mathbb{PR}^n , \mathbb{Z}/2)$ is given by

$a \otimes 1 \to a, 1 \otimes b \to 0$

and

$pr_n ^*:H^*(\mathbb{PR}^n , \mathbb{Z}/2) \to H^*(\mathbb{PR}^n \times \mathbb{PR}^m, \mathbb{Z}/2)$ by

$a \to a \otimes 1$.

My attempts:

$i_n^*, pr_m^*$ respect grades and therefore $pr_n^*$ maps $$a \to d_n a \otimes 1 + e_n 1 \otimes b$$

with unknown $d_n, e_n \in \mathbb{Z}/2$. Compose with $i_n^*$ gives $a = i^*_n \circ pr_n^*(a) = i^*_n(d_n a \otimes 1 + e_n 1 \otimes b) = d_n i^*_n(a \otimes 1) + e_n i^*_n(1 \otimes b)$ and $0 = i^*_m \circ pr_n^*(a) = i^*_n(d_n a \otimes 1 + e_n 1 \otimes b) = d_n i^*_n(a \otimes 1) +e_n i^*_n(1 \otimes b)$.

I get analogous system applying $pr_m ^*$ with another coefficients $d_m, e_m$. From here - without extra information - I'm stuck. Could anybody help?

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I think your misunderstanding stems from the fact that you are using the Künneth Theorem as an identification. You are writing $a\otimes b$ as an element of $H^*(\mathbb{R}P^m\times\mathbb{R}P^n;\mathbb{Z}_2)$, which it is not. Actually $a\otimes b\in H^*(\mathbb{R}P^m;\mathbb{Z}_2)\otimes_{\mathbb{Z}_2}H^*(\mathbb{R}P^n;\mathbb{Z}_2)$, so it does not make sense to apply the homomorphism $pr_1^*:H^*(\mathbb{R}P^m\otimes\mathbb{R}P^n;\mathbb{Z}_2)\rightarrow H^*(\mathbb{R}P^n;\mathbb{Z}_2)$ to this element, since it does not have the correct domain.

To see what is going on, you need to understand that the Künneth Isomorphism is induced by the cohomology cross product. For example, Hatcher, on page 214, defines the cross product (or external cup product) for cohomology over a ground ring $R$

$$K:H^*X\otimes_R H^*Y\rightarrow H^*(X\times Y)$$

by

$$a\otimes y\mapsto x\times y=pr_1^*x\cup pr_2^*y\in H^{m+n}(X\times Y),\qquad x\in H^mX,y\in H^nY,$$

where $X,Y$ are spaces and $pr_1:X\times Y\rightarrow Y$, $pr_2:X\times Y\rightarrow Y$ are the projections. A version of the Künneth Theorem appears on page 219 as Theorem 3.18 and states that the cross product $K$ is an isomorphism of rings if $H^*Y$ is a finitely generated free graded $R$-module. Certainly these conditions hold for $X=\mathbb{R}P^m$, $Y=\mathbb{R}P^m$ and $R=\mathbb{Z}_2$.

Thus you really apply $pr_1^*$ to the element $K(a\otimes b)=a\times b=pr_1^*a\cup pr_2^*b$. So as it turns out the second of your queries is tautologically true if you are willing to assume the Künneth formula as given, and that $i_n^*(a\times 1)=1$ then follows directly from your computations, since $(a\times 1)=pr_1^*a\cup pr_2^*1=pr_1^*a\cup 1=pr_1^*a$. Note the fact that $pr_1^*1=1$ follows from the fact the cross product is an isomorphism of rings.

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  • $\begingroup$ Ah yes, thank, I guess I see the core problem. One question: What about $i_n^*(a\times 1)=1$? Is this just an easy conclusion from property of $\cup$? By naturality of $\cup$? My idea would be $i_n^*(a\times 1)= i_n^*(pr_n^*a\cup pr_1^*a\cup 1)= (i_n^*pr_n^*a) \cup (i_n^*pr_m^*1 )= a \cup 0$ $\endgroup$ – KarlPeter Dec 9 '18 at 12:45
  • $\begingroup$ since $i_n^*pr_m^*=0$. Is it ok? The cruical point for me is if I can do $i_n^*(pr_n^*a\cup pr_1^*a\cup 1)= (i_n^*pr_n^*a) \cup (i_n^*pr_m^*1 )$ but I think that's ok since $\cup$ respects ring maps. I do I need another argument? $\endgroup$ – KarlPeter Dec 9 '18 at 12:45
  • $\begingroup$ Yes, it follows as you have guessed. But actually $i^*npr^*m=1$ is the ring unit, since it corresponds to the constant map $X\times Y\rightarrow \ast\rightarrow X\times Y$, and $\ast\rightarrow X\times Y$ induces an isomorphism on $H^0$ (assume path connected $X$, $Y$), which is where the class $1$ lives. $\endgroup$ – Tyrone Dec 9 '18 at 13:31
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    $\begingroup$ I'll write $i_i$, $i_2$ and $pr_1$, $pr_2$ for the inclusions and projections onto the first and second factors. Then I suppose what I mean is that the composite $pr_2\circ i_1:X\cong X\times \ast\rightarrow X\times Y\rightarrow \ast\times Y\cong Y$ is equal to the composite map $X\rightarrow \ast\rightarrow Y$ where $X\rightarrow\ast=\{pt\}$ is the map that crushes $X$ to a point and $\ast\rightarrow Y$ is the inclusion of the basepoint of $Y$. If and $X$, $Y$ are connected then it is easy to see that this latter map is an isomorphism on $\pi_0$, $H_0$ and more relevantly $H^0$. $\endgroup$ – Tyrone Dec 9 '18 at 14:04
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    $\begingroup$ To answer your question a little more directly, you do get $i_1^*(a\times 1)=i_1^*(pr_1^*a\cup pr_2^*1)=(i_1^*pr_1^*a)\cup(i_1^*pr_2^*1)=id_X^*a \cup 1=a$. Here $a\in H^nX$ and $1\in H^0X$ so that their cup product lives in $H^{n+0}X=H^nX$. On the other hand $i_2^*(a\times 1)=i_2^*(pr_1^*a\cup pr_2^*1)=(i_2^*pr_1^*a)\cup (i_2^*pr_2^*1)=0\cup 1=0$ since $pr_1\circ i_2$ factors $Y\rightarrow \ast \rightarrow X$, and $H^n(\ast)=0$. The difference is that $Y\rightarrow \ast\rightarrow X$ induces an isomorphism on $H^0$, but the zero map on $H^n$ for $n>0$. $\endgroup$ – Tyrone Dec 9 '18 at 14:21

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