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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?

savick01
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2 Answers2

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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 :)

mland
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  • Thanks! So, the candidate $c\in \mathrm{Map}(S^3,F_f)\subseteq \mathrm{Map}(S^3, {(\mathbb{C}P^\infty)}^I)\simeq \mathrm{Map}(S^3\times I, \mathbb{C}P^\infty)$ should be a homotopy between Hopf fibration and a constant function, right?

    The thing now is that we don't know Whitehead's and CW-complexes, so I'm expected to provide a solution without it (and of course it will be much easier for me at the moment). Can you think of an explicit homotopy inverse?

    P.S. What are the homotopy groups of $\mathbb{C}P^\infty$?

     – savick01 Jan 03 '13 at 12:48
  • @savick01 first the PS. There is a fibration $S^1 \to S^\infty \to \mathbb{C}P^\infty$ and $S^\infty$ is contractible. This tells you that $\mathbb{C}P^\infty$ is what is called a $K(\mathbb{Z},2)$ i.e. its homotopy groups are $\pi_i(\mathbb{C}P^\infty) = 0$ for $i \neq 2$ and $\pi_2(\mathbb{C}P^\infty) \cong \mathbb{Z}$. – mland Jan 03 '13 at 17:45
  • Now to your other question. I tend to think of a homotopy fiber only as a homotopy type, whereas you have a concrete model in mind. But to finish my idea of the proof I don't need the concrete model, all I need is that there is a long exact sequence in homotopy groups, and (well that's almost the same argument in the end) that the Hopf fibration tells me that higher homotopy groups of $S^3$ and $S^2$ are isomorphic via the Hopf map. – mland Jan 03 '13 at 17:49
  • At the moment I do not really have an elementary proof, in particular from the top of my head I do not have a map in mind that could serve as homotopy inverse. I understand that you want this and I will think about it, but let me nevertheless emphasize how cool the Hurewicz and Whitehead theorems are, because you do not need to have an inverse map, the information induced on homotopy groups suffices to determine the homotopy type... which I think is really cool. – mland Jan 03 '13 at 17:52
  • Great, I see now what I supposed from your proof about $\pi_n(\mathbb{C}P^\infty)$:) Yes, I understand your proof up to not being able to show $F_f$ has homotopy type of a CW-complex (and lack of experience with the Whitehead theorem). I was talking about a particular model because I hoped to find the inverse of $c$. But if it fails, a proof by Whitehead is much better than nothing. Actually, your proof gives me some new perspective and I would really enjoy writing it down as a homework.

    Would you be so kind to explain me why $F_f$ is a CW-complex?

     – savick01 Jan 03 '13 at 19:30
  • I'm not sure if you've noticed my question at the end of the previous comment, so I dare to ask again. For the Whitehead's we need to justify that $F_f$ is a CW-complex (up to a homotopy equivalence). Could you tell me why it is a CW-complex? Sorry if I'm asking stupid questions, but I have no experience with CW-complexes and theorems from my lecture notes doesn't help me here. – savick01 Jan 04 '13 at 22:20
  • I am still thinking about this. I had in mind that it is True, I will come back to you as soon as I have a good reason or a reference. I think it definitely involves the theorem that the loop space of a CW complex has a CW structure again.. – mland Jan 05 '13 at 11:51
  • So I now at least found a reference. There precisely your question is asked. Maybe in this pdf you will find further references. – mland Jan 05 '13 at 12:06
  • https://www.dpmms.cam.ac.uk/~bt219/fiber.pdf page 42, part 6 – mland Jan 05 '13 at 12:07
  • Thanks! In Introduction to Homotopy Theory by Paul Selick (bottom of page 73) it is said that a homotopy fibre of a map of CW-complexes is a CW-complex itself and it follows some Milnor theorem, but no reference is given (a reference is given where a special case is introduced, but I have no access to the reference part of the book). – savick01 Jan 05 '13 at 15:40
  • I found a paper of Milnor where in the introduction he writes that $(A,B,a_0)^{(I,{0},{1})}$ has CW-complex homotopy type which is exactly what we need here. But the whole thing is about 7 pages... – savick01 Jan 05 '13 at 15:44
  • Yes, I found that paper, too. I guess it is a nontrivial fact, that you may use the Whitehead theorem for this exercise. Moreover I still do not really know how to produce an inverse map $F_f \to S^3$. – mland Jan 05 '13 at 16:30
  • My lecturer showed me a more elementary way to prove this homotopial equivalence. I'll try to write it down soon.

    Coming back to your proof: I can't deduce all the necessary information about $\pi_n(F_f)$ for $n=1,2$ from the long exact sequence without showing manually that for example $\pi_1=0$. Am I missing something? Maybe it is known that $\pi_2(S^2)\to \pi_2(\mathbb{C}P^\infty)$ is an isomorphism?

     – savick01 Jan 06 '13 at 21:44
  • Yes, this is something you have to know as well. But it follows by cellular approximation since the 3-skeleton of $\mathbb{C}P^\infty$ is $S^2$. Curious to see your new proof :) – mland Jan 07 '13 at 08:19
  • Not sure if everything's OK, but I've just written it down. Enjoy:) – savick01 Jan 09 '13 at 20:02
2

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 $S^3\simeq \Omega BS^3$.

savick01
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  • In the Puppe sequence it should be a $\infty$ instead of $n$ right? I understand the proof modulo the fact that again I would prove the equivalence $S^3 \simeq \Omega BS^3$ using the Whitehead theorem. Because the way you state it don't you need that the homotopy inverse of $h$ is fiber-preserving? I don't really see this... – mland Jan 09 '13 at 21:30
  • Indeed, thanks. Yes, I need it. There is a theorem saying that for a homotopy equivalence that preserves fibres there is a homotopy inverse preserving fibres and homotopy functions between respective compositions and identity functions are fiber-preserving. I don't know any name behind this theorem but it was proved during my course. – savick01 Jan 10 '13 at 14:11
  • @savick01, do you have a reference for this theorem? – Sigur Jan 28 '15 at 23:13
  • @Sigur, it is written here as Twierdzenie 3.17 (in Polish).

    You can find a confirmation (but not the proof) at MO and in the Hatcher's book (remark below 4.61 left as excercise 3 in chapter 4.H).

     – savick01 Feb 01 '15 at 23:08
  • @savick01, thanks for the reference. It'll be a good chance to improve my pour Polish. – Sigur Feb 02 '15 at 11:41