[It is really annoying that the title cannot contain any tex code:( ]

Hi all, sorry for not having written much in the blog lately, i was kind of busy. First of all, I’m happy to announce that next I’ll be a graduate student at UT in Austin. I’m so happy and I’m looking forward to start working there. Ok, back to business, on today’s random exercise I will deal with a problem I encountered today. Since I found it funny and interesting I will write it down here. We are asked to study the homotopy fibre of the map

.

Let us call such homotopy fibre . First step is to compute its homotopy groups. A quick application of the l.e.s. of a fibration together with the fact that the inclusion induces an isomorphism in degree gives that:

Now we turn our attention to the integral cohomology of via a spectral sequence argument. God, I love spectral sequences 🙂

So the second page of the Serre Spectral sequence for the fibration looks like this

where in the upper rows I used the canonical isomorphism:

which can be obtained in several ways, UCT mainly. This is to stress the fact that we are going to make heavy use of the multiplicative structure of this SSeq.

**Claim 1: **The differential is uniquely determined by the differential

**Proof. **This is just an application of the multiplicativity of the SSeq together with the ring structure of the cohomology ring of . In fact we have that

but since it is a first quadrant spectral sequence. Therefore we see that

Proving the claim.

**Remark: **Since for , it should be clear to the reader that on the stable page the third, fourth, fifth and so on diagonals have to be trivial, i.e. whenever $p+q\geq 3$. It’s should be easy to understand why, if this is weren’t true, then we would obtain a non-trivial .

**Claim 2: **

**Proof. **before proving it, it’s clear that we made a little abuse of notation, what we really mean is that is an isomorphism. Let’s have a look at the third diagonal: the only possibly non zero object is . In order to ensure its triviality the only possibility (check it!) is that is injective. Now look at the fourth diagonal. In order to ensure the triviality of the only possibility is that $latex d_{0,3}^4$ is surjective. In fact it’s the only differential hitting it.

**Claim 3: **.

**Proof. **Looking at the fourth diagonal, it’s clear (by the same reasoning as above) that in order to ensure , has to be injective (again, it’s the only possibly non-trivial differential hitting/starting from position ). Since differential are, well, differential () together with **Claim 1** we have that for all . In particular being an injection it must be that as claimed.

**Claim 4: ** for

**Proof. **The reasoning now is the same as above, and is done inductively. I will show in detail the step for .

In the picture you can find that the only possible way to kill is by means of (the others, as shown, land in zeroes). Unfortunately, by **Claim 2 + Remark. **So the only possibility is that as claimed. The other cases are done similarly.

We can conclude that the homotopy fibre of is a Moore Space of type .

**Update #1:** I realised we can be more precise with the identification of the fibre: First of all, the map we are interested in, via cellular approximation can be homotope to the map . Then we use the (weak) homotopy equivalence between

Since is clearly a cofibration (CW inclusion), its homotopy cofibre is simply its cone, but the cone of $latex S^2 \hookrightarrow \mathbb{C}P^2$ is homotopy equivalent to , since we are simply “killing” the cell in dimension , leaving just the th and th cell. Now it’s easy to see that we have an homotopy equivalence (by Whitehead theorem applied to the map given by the usual adjunction). Since the homotopy fibre is defined up to homotopy, we can take as a model for it the -sphere (which is a Moore Space of type clearly )

**Update #2: **Actually I realised it’s way easier proving that . Recall we proved that and that . Take a representative of the generator of this group and call it . With this map (it’s important to have an *actual *map!) one can consider , and it’s really easy to realise that via the long exact sequence. Since we chose cleverly, it turns out that (again, by the l.e.s.). An application of the l.e.s. in homology for the pair together with Hurewicz gives that . Since is simply connected, by the relative Whitehead Theorem (**[FoFu]**, Lecture 14.5 page 183) we have that for all and is therefore a weak homotopy equivalence. Since the homotopy fiber is a CW complex, we have that is even an homotopy equivalence, and therefore we have the claim.

**[FoFu]** Fomenko, Fuchs, *Homotopical Topology – Second Edition*, Graduate Texts in Mathematics, Springer.