For which $n\in\mathbb{N}$ exists $f:S^{2n}\to \mathbb{C}P^n$ such that $f$ induces an isomorphism $H^{2n}(\mathbb{C}P^n)\to H^{2n}(S^{2n})$ My question is: For which $n\in\mathbb{N}$ exists a map $f:S^{2n}\to \mathbb{C}P^n$ such that $f$ induces an isomorphism $H^{2n}(\mathbb{C}P^n)\to H^{2n}(S^{2n})$ on singular cohomology with coefficients in $\mathbb{Z}$ (or cellular cohomology, depends on what suits better).
I already know that for $n=0$ such a map $f$ doesn't exist, because $S^0$ is a disjoint union of two points and $\mathbb{C}P^0$ is a point.  For $n>0$ I use Poincaré-duality, because it is $H^{2n}(\mathbb{C}P^n)\cong H^{2n}(S^{2n})\cong \mathbb{Z}$, and we can choose generator $\mu_1\in H^{2n}(\mathbb{C}P^n)$ and $\mu_2\in H^{2n}(S^{2n})\cong \mathbb{Z}$, such that $$\cap \mu_1:H^{2n}(\mathbb{C}P^n)\to H_0(\mathbb{C}P^n)$$ and $$\cap \mu_2:H^{2n}(S^{2n})\to H_0(S^{2n})$$ are isomorphisms (this is Poincaré-duality). Maybe one has to check now for which $f$ the diagramm \begin{align*}
 \require{AMScd}
 \begin{CD}
  H^{2n}(\mathbb{C}P^n) @>{H^{2n}(f)}>> H^{2n}(S^{2n})\\
  @V{\cap\mu_1}VV @VV{\cap\mu_2}V \\
  H_0(\mathbb{C}P^n) @>>{H_0(f)}> H_0(S^{2n})
 \end{CD}
\end{align*} commutes. But here I'm stuck. I appreciate your help, alternative ideas are welcome, too. 
Best.
 A: This is only possible for $n = 1$. When $n \ge 2$, $H^{2n}(\mathbb{CP}^n)$ is generated by $\alpha^n$ where $\alpha \in H^2(\mathbb{CP}^2)$, but since $H^2(S^{2n}) = 0$, $\alpha$ must map to zero in $H^{\bullet}(S^{2n})$, and hence so must $\alpha^n$.
A: E: Sorry, I misread the original post.
Let $M$ be a closed $n$-manifold. Then a degree $1$ map $S^{n} \to M$ is a homotopy equivalence.
Proof: First, $M$ is simply connected; for otherwise we could factor this map through $\tilde M$. If $\tilde M$ is noncompact then $H_n(\tilde M) = 0$, so the map is degree zero; if $\tilde M$ is compact then the cover is finite, say degree $d$, then because we have a factorization $S^{n} \to \tilde M \to M$, the map $S^n \to M$ must have degree divisible by $d$. 
Now suppose $M$ has a nontrivial cohomology class $\alpha \in H^k(M;\Bbb F)$ where $\Bbb F$ is either $\Bbb Q$ or $\Bbb Z/p$. (Because the cohomology groups of a compact manifold are finitely generated, if we show such things don't exist for all $0<k<n$, we show that $H^*(M;\Bbb Z) = H^*(S^n;\Bbb Z)$.) Then by Poincare duality the cup-pairing is nondegenerate so there is a class $\beta$ with $\alpha \smile \beta = 1 \in H^n(M;\Bbb F)$. (This is where we use the field coefficients.) So because $f^*: H^*(M;\Bbb F) \to H^*(S^n;\Bbb F)$ is a ring homomorphism, and because $f^*(\alpha \smile \beta)$ is $1$, we conclude $f^*(\alpha)$ is nonzero. Uh-oh - the cohomology of the sphere doesn't allow that. So there was no such $\alpha$ all along. What we've proved is that $M$ is simply connected and that $f: S^n \to M$ induces an isomorphism on cohomology. A version of the Whitehead theorem shows that $f$ is a homotopy equivalence, as desired. So in particular $M$ is not $\Bbb{CP}^n$. (Actually, because the Poincare conjecture is now known in all dimensions, $M$ is homeomorphic to $S^{2n}$.)
If you instead ask about maps $S^n \to M$ that have nonzero degree $d$, the above argument almost goes through; you may as well ask about simply connected manifolds; and the same argument shows that $M$ is a $\Bbb Z[1/d]$-homology sphere (as opposed to a $\Bbb Z$ homology sphere as above). Such simply connected things that aren't $S^n$ only begin to exist in dimension $5$.
In the other direction, to get a map $\Bbb{CP}^{n} \to S^n$, all you need is $n>0$. Pick an embedding of a closed disc $D^{2n} \hookrightarrow \Bbb{CP}^n$; then $\Bbb{CP}^n/(\Bbb{CP}^n \setminus \text{int} D^{2n}) \cong D^{2n}/(\partial D^{2n}) \cong S^{2n}$. So the quotient map gives a map $\Bbb{CP}^n \to S^{2n}$, and using the local degree formula as in Hatcher (which, if you look carefully at the proof, you'll see is valid for maps between any pair of closed $n$-manifolds) you see that this map is degree 1, as desired.
This construction works for any closed, connected $n$-manifold $M$ ($n>0$) to get a map $M \to S^n$ of degree 1. (Of course you can do it for non-closed $M$ but then degree doesn't make sense, and for non-connected $M$ but then you lose data about the components the disc isn't in). 
