Homotopy fiber of inclusion of projective spaces equivalent to sphere $S^3$ Consider the inclusion map $S^2=\mathbb{C}P^1 \overset{f}{\to} \mathbb{C}P^\infty$ ($\mathbb{C}P^\infty$ is the sum [direct limit] of $\mathbb{C}P^n$s) and the mapping space $E_f\subseteq \mathbb{C}P^1 \times {(\mathbb{C}P^\infty)}^I$ with a fibration $p_f: E_f \to \mathbb{C}P^\infty$ such that $p_f(c,\omega)=\omega(1)$. Prove that the fiber $F_f$ of $p_f$ (also known as the homotopy fiber of $f$) is homotopically equivalent with $S^3$.
In other words I want to prove that a family of paths (with compact-open topology) in $\mathbb{C}P^\infty$ starting at $\mathbb{C}P^1$ and ending at some fixed point (say, also in $\mathbb{C}P^1$) is equivalent to $S^3$.
I've thought about it a lot, but I have very little experience with projective spaces and no idea how to map $F_f$ to $S^3$ or how to somehow deform it to make it smaller and more similar to the sphere... Any hints?
 A: A hint could be the following:
You have a homotopy fiber sequence
$$ F_f \to S^2 \to \mathbb{C}P^\infty.$$
Look at the long exact homotopy sequence.
What can you deduce about $\pi_i(F_f)$ for $i=0,1,2,3$ ?
To get a candidate for a map between $S^3$ and $F_f$ that could be a homotopy equivalence look at the part of the long exact sequence that reads as:
$$ \dots \to \pi_4(\mathbb{C}P^\infty) \to \pi_3(F_f) \to \pi_3(S^2) \to \pi_3(\mathbb{C}P^\infty) \to \dots $$ 
Now try to show that this candidate induces an isomorphism on all homotopy groups, then apply Whiteheads theorem to conclude your claim.
If you have more questions I am happy to fill in details wherever you want me to :)
A: Another approach may be the following.
Let's look at the action of unit quaternions $S^3<\mathbb{H}^*\subseteq \mathbb R^4$ on $S^\infty \subseteq \bigoplus\mathbb R^4$. For the action of $S^1<S^3$ we have a fibration:
$$S^1\to S^\infty \to  S^\infty/S^1 =\mathbb CP^\infty$$
and for the action of $S^3$ we have:
$$(*)~~S^3\to S^\infty \overset{q}\to S^\infty/S^3 = BS^3.$$
It is not hard to notice that $S^3/S^1\to S^\infty/S^1 \to BS^3$ is also a fibration and the first map is the standard inclusion map. From the above we have a Puppe sequence:
$\Omega BS^3 \to S^2 \to \mathbb CP^\infty \to BS^3$
and all we need is to show that $\Omega BS^3 \simeq S^3$.
Let's have a look at another (different from $*$) fibration above $BS^3$: the space of paths $P(BS^3,p)$ in $BS^3$ starting at some point $p \in BS^3$ with the map evaluating paths at $1$: $P(BS^3,p)\overset{e_1}\to BS^3$. We can map $S^\infty $ to $P(S^\infty,p')$ using the fact that $S^\infty$ is contractible and further to $P(BS^3,p)$ by the quotient map $q$. Let's denote the whole compositon by $h$. We have the equality $q=e_1 \circ h$ and $h$ is homotopy equivalence (both spaces are contractible), so $h$ is a fiber homotopy equivalence, which proves that fibres of both fibrations are equivalent $S^3\simeq \Omega BS^3$.
