TITLE: Is the map $S^{2n+1}\rightarrow \mathbb{C}P^n \rightarrow \mathbb{C}P^n/\mathbb{C}P^{n-1}\cong S^{2n}$ essential?
QUESTION [15 upvotes]: This question is motivated solely by idle curiousity.
There is a natural map $p:S^{2n+1}\rightarrow \mathbb{C}P^n$ mapping a point on $S^{2n+1}\subseteq \mathbb{C}^{n+1}$ to the unique complex line it spans in $\mathbb{C}^{n+1}$.
It is well know that $p$ gives $S^{2n+1}$ the structure of an $S^1$ bundle over $\mathbb{C}P^n$.  From this, one gets a long exact sequence in homotopy groups.  Since $\pi_k(S^1)  = 0$ for $k > 1$, one sees from this that $p_\ast :\pi_{2n+1}(S^{2n+1})\rightarrow \pi_{2n+1}(\mathbb{C}P^n)$ is an isomorphism.  In particular, $p$ is homotopically nontrivial.
Likewise, notice that by collapsing the $2n-2$ skeleton of $\mathbb{C}P^n$ to a point, one gets a map $q:\mathbb{C}P^n\rightarrow \mathbb{C}P^n/\mathbb{C}P^{n-1}\cong S^{2n}$, the latter homeomorphism coming from the usual cellular picture of $\mathbb{C}P^n$.  It is easy to see that this map induces an isomorphism on $H_{2n}$, so is also homotopically nontrivial.
My question is about the composition $q\circ p : S^{2n+1}\rightarrow S^{2n}$.

Is the composition $q\circ p:S^{2n+1}\rightarrow S^{2n}$ also homotopically nontrivial?

Here are some data points:

The composition of two homotopically nontrivial maps need not be homotopically nontrivial.  The simplest examples are probably the following:  If $X$ and $Y$ are both noncontractible, then any of the "natural" inclusions $i:X\rightarrow X\times Y$ is homotopically nontrivial, the projection $X\times Y\rightarrow Y$ is homotopically nontrivial, but the composition is homotopically trivial.
In the case $n=1$, $q$ doesn't actually collapse anything - it's a homeomorphism.  Thus, in this case $q \circ p$ is homotopically nontrivial.  (In fact, it's known to generate $\pi_3(S^2)$).
For higher $n$, $\pi_{2n+1}(S^{2n})\cong \mathbb{Z}/2$, so the square of $q\circ p$ in $\pi_{2n+1}(S^{2n})$ is homotopically trivial.
If we use $\mathbb{R}P^n$ instead of $\mathbb{C}P^n$, we end up getting a map $S^n\rightarrow S^n$.  This map is homotopically nontrivial iff $n$ is odd, where it acts by multiplication by 2 on the top homology group.

Thanks, and please feel free to retag as appropriate!

REPLY [5 votes]: Here's another proof coming from Pontrjagin's framed cobordisms.
We'll use the following result due to Pontrjagin, which can be found, for example, in Milnor's Topology from the Differentiable viewpoint:

For any closed $n$-manifold $M$, the homotopy classes of maps into $S^k$, denoted $[M,S^k]$ is in 1-1 correspondence with framed (n-k)-submanifolds of $M$ (submanifolds with trivial normal bundle and a particular choice of trivialization) modulo the equivalence of framed cobordism (where the cobordism is required to be a submanifold of $M$ as well).

The correspondence is as follows:  given a homotopy class of map $[f]:M\rightarrow S^k$, choose a smooth representative $f$.  Choose any regular value $p$.  Then the associated submanifold is $f^{-1}(p)$ and the framing is pulled back from a choice of oriented basis for $T_p S^k$.
In our particular application, we want to understand a map from $S^{2n+1}$ to $S^{2n}$.  By dimension count, the inverse image will be a disjoint union of circles.  By drawing pictures in $\mathbb{R}^3$, one can easily convince oneself that any number of circles is frame cobordant to a single circle.
Our particular map is already smooth, at least, away from an equitorial $S^{2n-1}$ (and it can be homotoped to be smooth there as well).  Since the map from $\mathbb{C}P^n$ to $S^{2n}$ is generically $1-1$, this reduces the problem to considering the induced framing on any Hopf circle.
The normal bundle of the Hopf circle $(e^{i\theta}, 0,....,0)\in S^{2n+1} (\subseteq\mathbb{C}^{2n+2})$ is a $2n$-dimensional vector bundle.  Hence, it is determined by a choice of $S^1\rightarrow SO(2n)$.  Note that if two choices are homotopic, the corresponding framings are frame cobordant:  One can take a cylinder over $S^1$ where the $"t"$ coordinate of the cylinder corresponds to the time parameter in the homotopy.  This is clearly embeddable in $S^{2n+1}$.
We conclude that the map $S^{2n+1}\rightarrow S^{2n}$ is essential iff the corresponding map $S^1\rightarrow SO(2n)$ is.  It remains to compute that map.
To that end, let $v_i$ denote the vector in $\mathbb{C}^n$ whose only nonzero coordinate (in the $i$th slot), is $1$ and consider the basis $$\{v_2, iv_2, ..., v_n, iv_n$$ of $(i,0,...,0)^\bot \subseteq T_{(1,0...,0)} S^{2n+1}$.  It's easy to see that the framing at the point $(e^{i\theta},0,...,0)$ in the Hopf circle (which is pulled back from $\mathbb{C}P^n$ via the Hopf map) is given by the basis $$\{e^{i\theta}\,v_2, e^{i\theta}\,iv_2,...,e^{i\theta}\,v_n, e^{i\theta}\,iv_n\}.$$
In particular, in terms of this basis, the map $S^1\rightarrow SO(2n)$ maps $e^{i\theta}$ to the block diagonal matrix $$\operatorname{diag}(R(\theta), R(\theta), ..., R(\theta))$$ where $$R(\theta) = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin\theta & \cos\theta\end{bmatrix}$$ is the standard rotation matrix.
It is relatively well known that we can homotope such a map to one of the form $$\operatorname{diag}(R(n\theta), 1, ..., 1)$$ which is very well known to be homotopically essential iff $n$ is odd.