TITLE: Are there nontrivial continuous maps between complex projective spaces?
QUESTION [28 upvotes]: Are there maps $f: \Bbb{CP}^n \rightarrow \Bbb{CP}^m$, with $n>m$, that are not null-homotopic? In particular, is there some non-null-homotopic map $\Bbb{CP}^n \rightarrow S^2$ for $n>1$? Can we say anything more generally about the pairs $(n,m)$ that admit such maps?
Note that, in the real case, the answer to the first quesiton is yes and the second no: the Hopf map descends to a non-null-homotopic map $\Bbb{RP}^3 \rightarrow \Bbb{RP}^2$, and there's a natural bijection between $[\Bbb{RP}^n, S^1]$ and $H^1(\Bbb{RP}^n; \Bbb Z) = 0$.
This question asks about the $n \leq m$ situation.
Edit: Jason DeVito's post below answers both questions in the affirmative; there remains, though, the following interesting case:
Is there a map $f: \Bbb{CP}^2 \rightarrow \Bbb{CP}^1$ that is not null-homotopic?

REPLY [16 votes]: One person offering three answers to one post seems ludicrous (and this is a first for me), but I've really enjoyed this problem.

Theorem:  There is a homotopically nontrivial map $\mathbb{C}P^2\rightarrow S^2$.

(By my second post, this map must be unique up to homotopy.)
We will prove this by answering the question in post 2.  Namely, we show that given the two framed $S^1\subseteq \mathbb{R}^4\subseteq \mathbb{C}P^2$ (which I'll denote $S^1_1$ and $S^1_2$) are NOT framed cobordant in $\mathbb{C}P^2$.  More specifically, given a framed cobordism between the two $S_i^1$s, we show we can lift the whole picture to $S^5$.  However, since $\pi_5(S^4) = \mathbb{Z}/2\mathbb{Z}$, the Pontryagin-Thom construction will imply that there is no such cobordism in $S^5$.
So, assume for a contradiction that there is such a cobordism $W$ between the $S_1^2$ and $S_2^1$.  Since $W$ is framed, the normal bundle is trivial.  Let $i:W\rightarrow \mathbb{C}P^2$ denote the inclusion map.  Since $W$ is a surface with boundary, $H^2(W) = 0$, so the induced map $i^\ast:H^2(\mathbb{C}P^2)\rightarrow H^2(W)$ is trivial for trivial reasons.
By a small extension of Claim 1 from the second post, there is a lift $\tilde{i}:W\rightarrow S^5$, in the sense that $\pi\circ\tilde{i} = i$, where $\pi:S^5\rightarrow \mathbb{C}P^2$ is the canonical projection.
Note that the normal bundle of $\tilde{i}(W)$ is the pullback under $\pi$ of the normal bundle to $W$, so is trivial.  In fact, a choice of trivialization of the normal bundle of $W$ induces one of $\tilde{i}(W)$ in the following fashion.  First, since $\pi\circ\tilde{i} = i$, it follows that $\tilde{i}(W)$ is transverse to the Hopf fibers.  In particular, if $V\in T_p S^5$ with $p\in \tilde{i}(W)$ denotes the vector field tangent to the Hopf vectors, then the projection of $V$ onto $T_p \tilde{i} W$, denote $\rho(V)$ is not the identity, so $V - \rho(V)$ is non-zero and normal to $\tilde{W}$.
To trivialize the rest of the normal bundle over $\tilde{i}(W)$, we start by embedding the normal bundle of $W$ into $\mathbb{C}P^2$ as a tubular neighborhood of $W$ (and I'll call this map $i$ as well).  This tubular neighborhood deformation retracts onto $W$, so $\tilde{i}$ extends to this neighborhood, and gives an embedding of the tubular neighborhood into $S^5.$  Then one can use this extended $\tilde{i}$ to transport a choice of trivialization of $W$ to one of $\tilde{i}(W).$
In particular, we have a framing on $\tilde{i}(W)$ given by taking framing given by the previous paragraph together with $V - \rho(V)$ $\tilde{i}(W)$.  Thus, $\tilde{i}(W)$ is a framed cobordism between $\tilde{i}(S_1^1)$ and $\tilde{i}(S_2^1)$.
But, there is no such framed cobordism.  The idea is that since both $S^1_i$ are in a small chart in $\mathbb{C}P^2$, their lifts to $S^5$ are given by essentially the same formulas (in an appropriate chart on $S^5$), together with $V - \rho(V)$.  According to post 2, under the Pontryagin-Thom construction, these are representatives from the two equivalence classes corresponding to the fact that $\pi_5(S^4)$ has exactly two elements.
This contradiction implies $W$ can't exist.  So, we've found at least two framed submanifolds of $\mathbb{C}P^2$ which are not framed cobordant.  By the Pontryagin-Thom construction, this implies $[\mathbb{C}P^2,S^2]$ has at least 2 elements.  In particular, there a homotopically nontrival map from $\mathbb{C}P^2$ to $S^2$.
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If one traces through the Pontryagin-Thom construction, it follows that the following map $\mathbb{C}P^2\rightarrow S^2$ is homotopically nontrivial.  We first define it as a map into $S^3$, then compose with the Hopf map $S^3\rightarrow S^2$.
Consider the normal bundle $\nu$ of $S_2^1$, embedded into $\mathbb{C}P^2$, and once and for all, pick a point $p\in S^3$ and a basis $\{e_1,e_2, e_3\}$ of $T_p S^3$.  We map the entire $0$ section of $\nu$ to $p$.  Each fiber of $\nu$, isomorphic to $\mathbb{R}^3$, is mapped onto $S^3$ with all points sufficiently far from the origin being mapped to $-p$ (think stereographics projection) and so that the differential maps the basis $\{w_1,w_2,w_3\}$ from Post 2 to the basis $\{e_1,e_2,e_3\}$.  Finally, map points of $\mathbb{C}P^2$ wich are not in $\nu\subseteq \mathbb{C}P^2$ to $-p$.