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.

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Let me know if something is unclear, I will do my best to explain it!

Riccardo Pedrotti – Stable classification of four manifolds and signature related questions

**[Update #1 (4 April 2017)]** I **may **have completed the stable diffeomorphism classification for smooth oriented closed four manifolds with fundamental group the dihedral group. The two missing -types were and in the case where the in the Dihedral group was a multiple of . I have to triple-check what I’ve done so far, but the idea was to exploit the cofibre sequence given by James:

For vector bundles and over a space , with is the projection of the sphere bundle of . With such cofibre sequence and some algebraic observations I was able to compute the stable page of the relevant AHSS/JSS and therefore providing an answer to the missing cases. I am so excited! I will try to address now the *hard *part, provide some unstable result, before June.

I will update the pdf above as soon as I have some time, now I’m quite busy with the deadlines of the PhD programs I was admitted/ I’m on the waitlist

**[Update #2 (13 May 2017)] **I uploaded an updated version of the notes, with some minor corrections and updates on my results. A correction which is worth noticing is that it seems (according to my computations at least) that the secondary invariant is not a relevant stable diffeomorphism in our cases, since it is always possible to change the normal -type of an almost spin manifold in such a way that the sec-invariant is always trivial.

**[Update #3] **Sadly, I found an error in my computations. I’m trying to fix it, but for sure at least a case in the stable diffeomorphism classification has to be checked again! I’m really sorry for this but I try to be as much precise and sincere as possible, especially when dealing with math! I hope to be able to provide the complete classification asap!

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This is a quick exercise which I was asked to solve by a friend of mine and since I found it funny I decided to spell some more details about it in order to show the importance of the Bockstein exact sequence. This could be a typical exam exercise in “Topology 2” here in Bonn, so dear fellows students, read and comment if anything is unclear!

The following is exercise 7, page 366 from **Bredon’s Topology and Geometry**:

Show that there can be no -manifold with

Clearly one has to reason by absurd, so let us assume that there exists indeed such manifold . The absurd will be given by the Bockstein homomorphism.

First of all, a quick application of UCT gives us the following:

Let us denote with a generator of . Let a generator of .

**Claim 1** We can choose , where is the Bockstein homomorphism associated to the s.e.s.

**Proof: **It’s enough to prove that is an isomorphism. Let us have a look at the relevant piece of the exact sequences for

and

exactness gives us that is an isomorphism and naturality w.r.t. coefficients gives us that is an isomorphism as claimed.

**Claim 2** .

**Proof: **Start by noticing that , since we are working with field coefficients. Therefore we can consider (i.e. it exists) an element as the dual of . By Poincaré Duality we have that for or (but not !!), . Therefore the pairing:

proves that .

In order to finish the exercise, let us compute , recall that by assumption :

where we used the fact that the Bockstein for our s.e.s. is a derivation (you can find it, as an exercise on Bredon’s book. You can prove it by hand via diagram chasing of the definition).

**This is a clear absurd, which concludes our proof.**

Note that the exercise can be generalized to -dim Poincaré complexes with the same homology groups but , with any odd prime. Think about it, so you can see where we used what, which is always important to know.

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In particular, I want to prove that the Eilenberg MacLance space , for a *discrete *group is a model for the classifying space .

First of all, using l.e.s. this exercise is trivial: in fact we have the following fibration

gives us the long exact sequence of homotopy groups. Recall that and for ( is discrete). By definition is contractible, therefore the long exact sequences boils down to

for

This proves that is a model for the .

**what happens if we don’t know this? Ok, it’s an easy result everyone should know, but it’s good to have a proof which relies on a totally different approach, since it’s the perfect chance to revise some notions in covering theory.**

Our approach will be the following: given , we will show that it has the universal property of the classifying space . Let us start with the following *crucial *observation: begin discrete, a principal -bundle over is in particular a covering space of . As I said before, this is crucial and it’s a good sign that it is crucial otherwise we would be able to find counterexamples with the above proof. So, this is our situation so far:

we want to find a map which classifies . The first step to build such map is to observe that, *as a covering, * is classified by and such is automatically a normal covering of . In fact, by definition of principal bundle, we have a fibre-wise transitive -action on , but this is exactly the requirements to be a normal covering (see **[Hat] **page 70).

**Claim 1**

**Proof:** ** **first of all, the quotient we wrote above makes sense since we have shown a moment ago that was a normal covering of . This means that, by Prop. 1.39.(a) page 71 in **[Hat], **that $latex p_*\pi_1H$ is a normal subgroup of the fundamental group of . Again, by Prop. 1.39.(b), w have that the quotient is indeed .

**Claim 2 **Let be a connected CW complex and let be a . Then every homomorphism is induced by a map that it’s unique up to based homotopy.

**Proof: **This is Prop 1B.9 page 90 in **[Hat].** This result can be generalised but since we need just this I will stick with this weaker version.

Now by **Claim 1** we have the canonical quotient morphism which corresponds, thanks to **Claim 2**, to a map :

Notice that by construction

since is designed to kill . Therefore there exists a lift (Prop 1.33 page 61 in **[Hat]** ) which we call making the diagram commute:

Now the crucial observation is that is a -map. This is a consequence of the commutative square above and the definition (via lifting of loops) of the action on both of the total spaces.

**Claim 3 **We have

**Proof: **After observing that is an isomorphism on fibres (consequence of being a -map) one sees that fulfils the universal property of the pullback bundle, and therefore we have the claim.

**Claim 4. **Isomorphic principal -bundles gives rise to homotopic maps and homotopic maps gives rise to isomorphic bundles.

**Proof: **The second assertion is easily seen to be true, thanks to the fact that homotopic maps gives rise to isomorphic pullback bundles. For the first half, let . Being isomorphic, we have by prop 1.37 page 67 in [Hat] that , therefore the *same map* can be used to build $u$ which is *unique up to homotopy.*

This concludes the exercise, since for every (isomorphism class of) principal -bundle we construct a (homotopy class of) map classifying it.

**Reference:**

**[Hat]** A. Hatcher, *Algebraic Topology*, Cambridge University Press.

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Since we will make an extensive use of this tool, it’s important that we clarify some technicalities about it. Our first aim is to prove the following lemma, which can be found as Lemma 2.3.2. page 27 in **[Tei92]**

**Lemma 1**

Let be a spectrum and the Atiyah-Hirzebruch spectral sequence as above.

- The differential is the dual of
- The differential is reduction mod composed with the dual of .

We start with a definition:

**Definition 1.** An homology operation is a natural transformation between homology functors.

We call them *stable* if they commute with the suspension isomorphism, in analogy with the well-known stable cohomology operations. Now we want to relate this operations with the more famous cohomology operations, and we will do it as follows:

**Lemma 2.** Let be two spectra. Let be the respective homology and cohomology theories . There is a bijection between stable homology operations and stable cohomology operations for .

**Proof.** By Yoneda lemma, we know that stable cohomology operations are represented by (homotopy class of) maps of spectra . Therefore given , we can define the following natural transformation:

Since we used a map of spectra, the natural transformation preserves the suspension isomorphism (See **[Rud]** page 69) and it represents a stable homology operation by definition. Now let be a stable homology operation. Since is determined by its value on finite spectra (just use a colimit argument), it’s enough to consider just the case where is a finite CW-spectrum. We can apply the Spanier-Whitehead duality to pass from cohomology to homology, and we obtain:

Therefore we created a cohomology operation, and after checking carefully the construction of such duality, it’s clear that these two assignations are one inverse of the other.

Now we want to give a construction of the AHSS in such a way that the differentials are clearly induced by maps of spectra. This is done in the paper **[Mau]** of Maunder, where the author build a version of the AHSS via an exact couple given by the Postnikov tower of the (co)homology theory. It’s clear that the construction done there (for spaces) can be lifted to the setting of spectra. He then identifies the differentials with -invariants of the spectrum representing the (co)homology theory. We can say something similar for the edge homomorphisms too, in fact they are defined from the map in the definition of the exact couple. By the work of Maunder, all of these maps respect the suspension isomorphism since they come from maps of spectra (See here for a long argument or just use Yoneda Lemma).

Recall this result.

**Theorem.** *Serre’s Theorem on the cohomology of .* For we quote the results obtained by Serre in **[Ser]**. In each of the following, we describe the algebras over the Stennrod algebra.

- Let be the mod reduction of the fundamental class,

We are ready now to prove **Lemma 1**

**Proof.** Using the inclusion of the bottom cell , which induces an isomorphism in homotopy in degree , we can compute the differentials for the spectral sequence:

In fact we are interested in differentials which start at most at the second row, since the third is trivial thanks to . Now the differentials are stable homology operations and thus we have the following cases:

- . In this case, the homology operation is classified by elements of , where is the Eilenberg-MacLane spectrum which represents singular homology with coefficients . By definition, . In fact the Steenrod algebra is well understood, and we know that the cohomology group we are interested in is generated by the Steenrod square . Now recall the identification:, where denotes the sphere spectrum. We can consider the following commutative diagram given by the Spanier-Whitehead Duality: Where is the map representing the stable homology operation . Now notice that we have the pairing given by where is the duality and the multiplication given by the ring spectrum structure. This is (under the duality isomorphism) the well-known Kronecker pairing which in the case of field coefficient is a perfect pairing. For this reason we can identify with the dual of , and becomes the dual of a stable cohomology operation . After plugging in some test-space (as done in
**[Tei92]**), one realises that it’s not the trivial operation, therefore the claim. - . Here the reasoning is a little bit more involved. First of all, one has to realise that now we are interested in elements of , which by definition is . Therefore we need to compute this last cohomology group. Using the isomorphism we need to figure out how does the look like for big enough . By Serre’s Theorem , we have that , where is the fundamental class. It’s easy to see that the colimit is given by the equivalence class of this element, therefore we have only two possible homotopy classes of map of spectra, the trivial one, and the one induced by . As above, by a careful choice of test space, one can see that can’t be always trivial, therefore it remains to see what’s the effect of on our homology groups in order to be able to compute it explicitly. To this end, we consider the following diagram: Now notice that . By definition, an element can be represented by a map . The effect of composition with it’s the well-known reduction modulo in homology. After that, by the same reasoning above, the effect of composition with is given by the dual of the Steenrod square on the cohomology of .

We can give a nice description of the horizontal edge homomorphism in the AHSS for Oriented and Spin Bordism. Actually this can be generalised to other bordism theories, but we are interested in these two.

**Definition 2.** Let be or . We define the Steenrod-Thom map for -bordism as follows:

This map is clearly natural.

We start by recalling the geometric interpretation of the -bordism homology in the relative case.

**Lemma 3.** Let be a CW-pair. We can identify with the abelian group of singular manifolds (possibly with boundary) with spin structure, i.e. it’s elements are -bordism classes of manifolds with maps . The boundary operator turns out to be equal to the map .

**Proof.** See the observation in **[Rud]** page 289.

The Steenrod-Thom map can be easily generalized to a map as follows:

Notice that the following diagram commutes

Since . Therefore the family of maps represents an homology operation and by **Lemma 2** is induced by a morphism of spectra .

We are now ready to prove the second main result (Prop 7.23 page 292 in **[Rud]**)

**Proposition.** The edge homomorphism:

is given by the Steenrod-Thom homomorphism.

**Proof.** First of all, notice that the edge homomorphism is a stable homology operation. In fact if we use the exact couple given in **[Mau]** to define the AHSS, every map involved is clearly a stable homology operation. Therefore it has to be represented by an element in , where is the Thom class of the spectrum . Moreover, similarly to what we did for the differential, we see that the edge homomorphism acts by post-composition of for some :

Since clearly generates such cohomology group, by Proposition 5.24.i page 260 in **[Rud]** we have that is an isomorphism. We choose such that . After plugging into the AHSS some test spaces, it’s clear that the edge homomorphism is represented by and not by any of its multiples. Now it remains to be shown that has the same effect on homology as the Steenrod-Thom map. To this end, consider the map of spectra . Since is an isomorphism the element must be a generator. So . But both generators maps and therefore

.

**Bibliography:**

**[Tei92], **P. Teichner, *Topological four-Manifolds with Finite Fundamental Group*, PhD thesis

**[Rud], **Rudyak, *On Thom spectra, Orientability, and Cobordism*, Corrected printing 2008, Springer Monographs in Mathematics

**[Mau], **C.R.F. Maunder, *The spectral sequence of an extraordinary cohomology theory*, Mathematical Proceedings of the Cambridge Philosophical Society 59, 1963

**[Ser], **J.-P. Serre, *Cohomologie modulo 2 des complexes d’Eilenberg-Maclane,* Comment. Math. Helv. 27 (1953), 198-232.

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There might be some errors here and there, please read with cautions and let me know if you find something unclear!

pedrotti-riccardo-cohomology-of-groups

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Ok, back to Math As the title suggests, I want to write something about this known result, since I needed it for understanding *why the first non-vanishing differential in the Atiyah-Hirzebruch Spectral Sequence is a stable operation. *What I’m going to show is not the nicest way to prove it, but still I think it’s worth to take a look at a more “concrete” proof

First of all it’s better to revise some concepts: recall the concept of the fundamental class for a E-M space . You can find something here at the bottom of the article.

Let be the canonical structure map for the E-M spectrum representing . In other words, $\rho_{A,n} \colon \Sigma K(A,n)\to K(A,n+1)$ is the inclusion. (To convince yourself just use a little bit of cellular approximation and Freudenthal Suspension Theorem)

Now we need to introduce some version of the suspension isomorphism in cohomology.

**Suspension Isomorphism 1: **Let us denote with

the map sending a class to , where is a generator. In Tammo tom Dieck’s Algebraic Topology, page 414, Proposition 17.3.1 it’s proven that , where is tom Dieck’s suspension defined at page 249, 10.2.5. Therefore we have $\Sigma^*=\sigma$.

**Suspension Isomorphism 2: **In Harper’s Secondary Cohomology Operation, at page 22 there is defined the so called Mayer-Vietoris suspension:

as the following composition:

where and are the inclusions and are the lower and upper cone of respectively.

**Claim 1: ** Given , then .

**Proof: I**t’s just an application of the naturality of the l.e.s of the pair and of the respective triple, as indicate in this diagram:

where starting at the lower left position and going clockwise gives you , going counter-clockwise gives you $latex (\Sigma f)^*\circ \Delta^*$.

**Claim 2:**

**Proof:** ** (Idea) **It’s enough to show that . A little bit of diagram chasing and naturality of the boundary operators for the map which pinch the lower extremity of the cone should do the trick. We make use of the definition of the boundary operator for the l.e.s. of the triple in cohomology (have a look here for the homology one, (we are looking at the second triangle pointing down in the top row)

We are almost ready to tackle down our original problem. We need the following technical lemmas. Let be the map of the UCT, and let the Hurewicz isomorphism from homotopy to homology.

**Claim 3: **The following diagram commutes for an appropriate choice of the polarising isomorphism:

**Proof: **The upper triangle commutes since Hurewicz map can be made compatible with the boundary operator and therefore the suspensions (Have a look at Harper’s Secondary Cohomology Operator page 24 in the middle). The lower trapezoid commutes because we can inductively define to do so, since is an isomorphism in degree (use cellular approximation).

**Claim 4: **We have

**Proof: **This is the diagram we want to prove commutativity:

which boils down to show that the two elements in the lower right corner in this diagram are the same:

but this is true since we know that every element is of the form for some . Using it we are done.

**Claim 5: **With the above notation, we have

**Proof: **Now we start to make use of the results done so far. We have the following equalities:

where we sued the definition of the fundamental class. Now we use naturality of UCT and Hurewicz to get

By **Claim 3 **we have

which is equal to

Now observe that, after composing with $\Phi$ both sides, it’s equivalent to prove that

but using **Claim 4 **we can move the suspension isomorphism inside the first UCT to the right:

Therefore we have the conclusion.

We now return to the stable cohomology operations stuff (finally).

**Claim 6 ** is stable cohomology operation if and only if

**Proof:** It’s easy to see that is stable cohomology operation if and only if it commutes with the suspension isomorphism. Now we have the following chain of equivalencies:

where we used naturality and **Claim 5.** This shows that the two conditions in the thesis are equivalent as claimed.

Now we are done for good.

**Theorem: **Let now be a map of spectra between (shifted) E-M spectra, then in cohomology, the components ‘s build together in a stable cohomology operation.

**Proof: **By definition of map of spectra we have the following commutative diagram:

which gives the following relation:

Now using **Claim 5 **and **Claim 1 **we have that

Since we proved that by **Claim 6 **we conclude that is a stable cohomology operation as claimed

The more theoretical approach would be to prove that the cohomology functor is *representable *and therefore natural transformations between representable functors are in natural bijection with appropriate maps between the *representing objects i.e. spectra.*

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The homotopy fibration induces an exact sequence in cohomology

I managed to find a solution to it now, and I thought it could be a good idea to write it here:

So let us start with a definition:

**Def 1. ** is an homotopy fibration sequence if a homotopy is given from the composed map to a constant map $latex z$ and the resulting map from $latex X$ to the homotopy fiber of $latex Y \to Z $ over is a weak homotopy equivalence.

So we start by proving that the sequence we have is really an homotopy fibration.

**Claim 1. **The sequence is an homotopy fibration sequence.

**Proof: **Consider the following square:

where we used the fact that the universal cover of the Eilenberg-Maclane space is a model for . The nullhomotopy for is provided by the contractibility of . Now consider this pullback diagram:

where the bended arrow is provided by the null-homotopy chosen before. Notice that that fiber of is really the homotopy fiber of since we can assume that is a fibration. Moreover, since monomorphisms are stable under pullback, the map is really the usual inclusion of fibre. A quick application of the l.e.s. of homotopy groups for a Serre fibration, gives us that

(use the fact that induces iso on first homotopy group). By the above diagram should be clear that the induced map is a weak equivalence. Therefore we concluded the proof of the fact that the sequence is an homotopy fibration.

**Prop 1. **Let be a fibration with path connected and based. Set . Assume is -connected and is -connected. Then there’s a exact sequence

we will prove this proposition in several steps.

Recall first the following theorem, called the Dual Blakers-Massey Theorem for squares [Munson & Volić, Cubical Homotopy Theory, Thm 4.2.2. page 188]

**Theorem 1. **Suppose that the following square

is an homotopy pullback square, and that the maps and are respectively -connected for . Then the canonical map is connected.

**Claim 2. **In the above setting,** **the map is -connected.

**Proof. **Consider the following square

we want to apply Theorem 1 to it, therefore we need to show that the square is homotopy pullback (see def. 3.2.4 page 102 in Cubical Homotopy Theory). By definition the homotopy pullback is the subspace of such that is s.t. and , in particular it’s the subspace consisting of points where . If we denote with the subspace of whose paths end in , we have that the homotopy pullback is just . Now it’s well-known that deformation retract to the constant path , therefore, by definition , proving that the square is indeed homotopy pullback. Notice then that the map is -connected since the cone is contractible and[ is -connected by assumption.

Therefore by Theorem 1 we have that the map is connected. Following definition 3.6.3 page 138 of Cubical Homotopy Theory, it’s immediate to see that which concludes the proof.

**Proof of Prop 1: ** Consider the cofibration sequence . Consider the Puppe Sequence for for large enough. Now use the fact that a equivalence gives us isomorphism in cohomology up to [tom Dieck, Algebraic Topology. Theorem 9.5.2 page 236] (so to conclude.

Back to our initial problem: our setting was , and , since . After applying our machinery we have the result. Notice that and therefore using UCT for cohomology it’s even in coefficient.

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**Proposition: **Every -homology class of a smooth 4-manifold is generated by a surface

*The classical proof goes as follows*

**Proof: ***(from Kirby’s Topology of -Manifolds Theorem 1.1 page 20)* There is an isomorphism:

so letting being the Poincaré dual of a chosen , there is an homotopy class of maps corresponding to . By cellular approximation, we can homotopy (a representative of) in order to obtain a map . In fact the 4-skeleton of is and cellular approximation tells you that the image of lies there. Make smoothly transverse to . Consider , *this will be an oriented surface representing .*

**yes but… why **That’s the real question, why represents our homology class

we need the following facts:

**Claim 1: **Let be a submanifold and let the homology class it defines in . Then in the Thom class of the normal bundle is dual to .

**Proof: **Have a look here .

Back to our original exercise. First of all we need to fix some notations: Let us denote with the Thom Class associated to the normal bundle of the inclusion . Similarly, let be the Thom class associated to the normal bundle of . With we will mention the Poincaré Dual of (both in the case of homology and cohomology).

So let be our homology class, as we said before , where is a map as above and is the fundamental class associated to ( or in our case, doesn’t change anything by cellular approximation). Using the fact that (see the addendum below), we have:

By **Claim 1 **and naturality of the Thom classes (the normal bundle over is the pullback via of the one over )

Using PD again,

Using **Claim 1 **again,

Which was exactly what we needed to show.

**ADDENDUM: **Let us recall briefly the definition of the fundamental class. Let be a *polarised *Eilenberg-MacLane space of type (i.e. we fixed an isomorphism from the homotopy group to . Using UCT (with map ) there is a unique class , called *the fundamental class *s.t. the composite

is the chosen . (it depends on the chosen iso!). In fact is connected, and abelian, the Hurewicz map is an isomorphism and therefore

Is a group homomorphism (an iso actually). Since UCT in degree is an iso, we define .

In our case, , and since , the map is a generator of it, and UCT maps generator to generator. So up to a sign, we can assume .

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Let and be compact oriented manifolds with smooth embedding . Let . Show that the Poincare duality isomorphism maps the cohomology class dual to to the homology class . Assume moreover that the normal bundle is oriented so that is orientation preserving isomorphic to .

**Hint: **consider the following commutative diagram:

where is a tubular neighbourhood of in .

Before going all in with the proof, let me recall two things:

**Theorem [Corollary 11.2 page 117 in Milnor’s Characteristic Classes] ***If is embedded as a closed subset of , the cohomology ring associated with the normal bundle in is canonically isomorphic to the cohomology ring . Here can be any coefficient ring.*

**Definition: ***Let be oriented, let be its Thom Class (denoted occasionally also ) Define to be the image of the Thom Class under the above iso. *

**Theorem [Thm 11.3 page 119]: **

*with coefficients, maps the fundamental class to the top Stifle-Whitney class of the normal bundle. Similarly, if is oriented, then the corresponding composition with integer coefficients maps the integral fundamental class to the Euler Class .*

**Definition: ***The image of is called the dual cohomology class to the submanifold of codimension . The terminology will be explained in the exercise*

**Proof of exercise: ** Denote with the fundamental class of .** **Start picking in the upper left angle the element . We need to chase this element a little bit. The left “path” ( and then ) is straightforward. We end up with the element .

For the “right” path ( and then and then ) we can use the so called coherency of the fundamental class and Corollary 11.2 to prove that is sent to which is the generator of ( See Bredon’s Topology and Geometry Theorem 7.8 page 344). Following the next arrow, which is an iso, we end up in with . *For the curious reader (yes please!) interested in this kind of stuff, I worked with compactly supported sections of the orientation bundle of , which we know being trivial since is orientable, for more details see Bredon’s Topology and Geometry chapter VI.7.*

Since the vertical right arrow (after choosing the Thom class of the normal bundle as the cohomology class in ) is the so called *homological Thom isomorphism* (tom Dieck’s Algebraic Topology Theorem 18.1.2 page 439), it’s an isomorphism so it maps to (up to a sign and using ). The conclusion follows at once.

(inspired by my answer here in M.Se)

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