What does this map $\mathbb{R}P^\infty\to\mathbb{C}P^\infty$ induce on cohomology? I am learning about algebraic topology, and I want to figure out what the map $f:\mathbb{R}P^\infty\to\mathbb{C}P^\infty$ defined by
$$f(\langle a_0,a_1,\ldots\rangle)=\langle a_0+0i,a_1+0i,\ldots\rangle$$
induces on cohomology. 
Here is what I have been thinking. The problem  is that this map is not a cellular map, since it sends the $k$-cell of $\mathbb{R}P^\infty$ to the $2k$-cell of $\mathbb{C}P^\infty$. I know that $f$ must be homotopic to a cellular map, by cellular approximation theorem, but I don't know how to find that explicitly. I also know that $H^*(\mathbb{R}P^\infty)\cong\mathbb{Z}[x]$ where $|x|=1$ and that $H^*(\mathbb{C}P^\infty)\cong\mathbb{Z}[y]$ where $|y|=2$. 
I guess the map $f^*:H^*(\mathbb{C}P^\infty)\to H^*(\mathbb{R}P^\infty)$ is just $\mathbb{Z}[y]\to\mathbb{Z}[x]$ with $y\mapsto x^2$, but I don't know how to show it. Could anyone help me please? Thanks.
 A: First of all, your formula for the cohomology of $\mathbb{R}P^\infty$ is wrong (I think you may have gotten confused with mod $2$ cohomology).  The correct formula is $H^*(\mathbb{R}P^\infty)\cong\mathbb{Z}[t]/(2t)$ where $|t|=2$.  In any case, the induced map $f^*$ on cohomology is determined by $f^*(y)$, since it must be a ring homomorphism.
Here is one way to compute $f^*(y)$.  First, note that the inclusion map $i:\mathbb{R}P^2\to\mathbb{R}P^\infty$ induces an isomorphism on $H^2$, so actually we can just compute $i^*f^*(y)$.
Now let's think about the map $fi:\mathbb{R}P^2\to\mathbb{C}P^\infty$. This map sends $[x,y,z]$ to $[x,y,z,0,0,\dots]$.  Note that this is homotopic to the map that sends $[x,y,z]$ to $[x+iy,z,0,0,\dots]$ (first linearly interpolate from $[x,y,z,\dots]$ to $[x+iy,0,z,\dots]$ and then linearly interpolate to $[x+iy,z,0,\dots]$).  Let us call this map $jg$, where $j:\mathbb{C}P^1\to\mathbb{C}P^\infty$ is the inclusion and $g:\mathbb{R}P^2\to\mathbb{C}P^1$ sends $[x,y,z]$ to $[x+iy,z]$.  Since $j$ induces an isomorphism on $H^2$, it is enough to compute what the map $g$ does on $H^2$.
But the map $g$ is very easy to understand geometrically.  This map $g$ is surjective and is injective except that it sends all points $[x,y,0]$ to $[1,0]$.  This means that $g$ is just the quotient map $\mathbb{R}P^2\to S^2\cong\mathbb{C}P^1$ that collapses the $1$-skeleton of $\mathbb{R}P^2$ to a point.  In particular, $g$ is cellular, and we can easily compute via cellular cohomology that $g^*:H^2(\mathbb{C}P^1)\to H^2(\mathbb{R}P^2)$ is nontrivial, and more specifically maps $j^*(y)$ to $i^*(t)$.  (Here $H^2(\mathbb{C}P^1)\cong\mathbb{Z}$ is generated by $j^*(y)$ and $H^2(\mathbb{R}P^2)\cong\mathbb{Z}/2$ is generated by $i^*(t)$.)
Now we can unwind all the isomorphisms from above to find $f^*(y)$.  We have $g^*j^*(y)=i^*(t)$.  But $jg$ is homotopic to $fi$, so $i^*f^*(y)=g^*j^*(y)=i^*(t)$.  Since $i^*$ is an isomorphism, this implies $f^*(y)=t$.
(Actually, in this case, since the only options are $f^*(y)=0$ and $f^*(y)=t$, it is enough to observe that since $g^*$ was nontrivial, $f^*$ must be nontrivial as well, so the only possibility is $f^*(y)=t$.)
