## Submersion between $S^{2n+1}$ and $\mathbb{C}P^n$ [closed]

How construct a submersion $\Lambda:S^{2n+1}\to\mathbb{C}P^n$ such that for every $x\in\mathbb{C}P^n$ , the fiber $\Lambda^{−1}(x)$ is diffeomorphic to the circle $S^1$ ?

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This is simply the restriction of the defining submersion $\mathbb{C}^{n+1}\setminus\lbrace0\rbrace \to \mathbb{C}P^n$ and is explained in any place where the projective space is defined for the first time. It's hardly at an appropriate level for MO and I'm voting to close. – José Figueroa-O'Farrill Sep 7 2010 at 16:46
For example, see the section "construction" here: en.wikipedia.org/wiki/Complex_projective_space – Ryan Budney Sep 7 2010 at 17:03
I wouldn't mind someone telling me if this can be done in more than one way up to some sensible equivalence... – Mariano Suárez-Alvarez Sep 7 2010 at 17:08
Bundles over $\mathbb CP^n$ with fiber $S^1$ are classified by homotopy-classes of maps $\mathbb CP^n \to BSO_2$, which is $H^2 \mathbb CP^n$. And you can check only two such bundles have total space $S^{2n+1}$. That's not what you meant, is it? – Ryan Budney Sep 7 2010 at 17:13
To elaborate a little on Jose's comment: Perhaps you are reading a reference where $\mathbb{CP}^n$ is defined as $\mathbb{C}^{n+1} \setminus \{ 0 \}/\mathbb{C}^*$. Then the $S^{2n+1}$ can be seen as the set of $(z_0, z_1, \ldots, z_n)$ in $\mathbb{C}^{n+1}$ such that $\sum |z_i|^2 = 1$, and the $S^1$ is the subgroup of norm $1$ elements in $\mathbb{C}^*$. – David Speyer Sep 7 2010 at 17:44