I am trying to understand what an axial map $f \colon \mathbb{R}P^n \times \mathbb{R}P^m \to \mathbb{R}P^k$ does on cohomology, where a map is called axial, if the restrictions $* \times \mathbb{R}P^m \to \mathbb{R}P^k$ and $ \mathbb{R}P^n \times * \to \mathbb{R}P^k$ induce isomorphisms in $\mathbb{F}_2$-cohomology. For this reason it is necessary, what the inclusion $i \colon \mathbb{R}P^n \to \mathbb{R}P^n \times \mathbb{R}P^m$ for some choice of a basepoint does, and since the inclusion composed with the projection is the identity, it is helpful to understand the cohomology of the projection. Now to my question:
Suppose we have two spaces $X = Y = \mathbb{R}P^{\infty}$ and the projection $\pi_1 \colon X \times Y \to X$. Let $F = \mathbb{F}_2$ be the field with two elements. It is $H^1(\mathbb{R}P^{\infty};F) \cong F$, and by the Künneth-theorem follows $H^1(X \times Y; F) \cong H^0(X;F) \otimes H^1(Y;F) \oplus H^1(X;F) \otimes H^0(Y;F) \cong F \otimes F \oplus F \otimes F$.
We have now the projection $\pi_1 \colon X \times Y \to X$ which induces the map $\pi^* \colon H^1(X; F) \to H^1(X \times Y;F)$. Let $x \in H^1(X; F)$ and $y \in H^1(Y;F)$ be the generators. Since $H^1(X \times Y;F) \cong H^0(X;F) \otimes H^1(Y;F) \oplus H^1(X;F) \otimes H^0(Y;F)$, a basis for $H^1(X \times Y;F)$ as a $F$-vector space is given by $1 \otimes y, x\otimes 1$, i.e. the elements of $H^1(X \times Y;F)$ have the form $a(1 \otimes y) + b(x \otimes 1)$ for $a, b \in F$.
I want now show, that $\pi_1^*(x) = x \otimes 1$ (after applying the isomorphism from the Künneth-theorem). Does anybody know how to do this?
The Künneth Theorem doesn't just state that they are isomorphic, it states that a specific map between them is an isomorphism.
There is a natural map $\Phi : H^*(X; F)\otimes H^*(Y; F) \to H^*(X\times Y; F)$ which is determined by $x\otimes y \mapsto \pi_1^*x\cup\pi_2^*y$ where $\pi_1 : X\times Y \to X$ and $\pi_2 : X\times Y \to Y$ are the projections. The element $\pi_1^*x\cup\pi_2^*y$ is sometimes denoted $x\times y$ and is referred to as the cross product or external cup product of $x$ and $y$.
Künneth Theorem: If $H^k(Y; F)$ is a finite-dimensional $F$-vector space for every $k$, then $\Phi : H^*(X; F)\otimes H^*(Y; F) \to H^*(X\times Y; F)$ is an isomorphism of graded rings.
See Theorem $3.15$ of Hatcher's Algebraic Topology for example - there the theorem is stated more generally with ring coefficients and $H^k(Y; R)$ is required to be a finitely generated free $R$-module, but every module over a field is free.
In particular, we get an isomorphism $\Phi : \bigoplus\limits_{i+j = m}H^i(X; F)\otimes H^j(Y; F) \to H^m(X\times Y; F)$ for each $m$.
When $m = 1$, we have $\Phi : H^1(X; F)\otimes H^0(Y; F)\oplus H^0(X; F)\otimes H^1(Y; F) \to H^1(X\times Y; F)$. If $Y$ is path-connected, then $H^0(Y; F) \cong F$ and we can write $1$ to denote a choice of generator. Now note that $\Phi(x\otimes 1) = \pi_1^*x\cup\pi_2^*1 = \pi_1^*x\cup 1 = \pi_1^*x$; that is, the element $\pi_1^*x \in H^1(X\times Y; F)$ corresponds to $x\otimes 1 \in H^1(X;F)\otimes H^0(Y;F)$ under the isomorphism $\Phi$. Likewise, if $X$ is path-connected, the element $\pi_2^*y \in H^1(X\times Y; F)$ corresponds to $1\otimes y \in H^0(X;F)\otimes H^1(Y;F)$ under the isomorphism $\Phi$. Therefore, if $X$ and $Y$ are path-connected, every element of $H^1(X\times Y; F)$ is of the form $\Phi(x\otimes 1 + 1\otimes y)$.
The same arguments can be used to show that if $Y$ is path-connected and $x \in H^i(X; F)$, then $\pi_1^*x \in H^i(X\times Y; F)$ corresponds to $x\otimes 1 \in H^i(X; F)\otimes H^0(Y; F)$ under the isomorphism $\Phi$. Likewise, if $X$ is path connected and $y \in H^i(Y; F)$, then $\pi_2^*y \in H^i(X\times Y; F)$ corresponds to $1\otimes y \in H^0(X;F)\otimes H^i(Y; F)$ under the isomorphism $\Phi$.