MAS61015 Algebraic Topology

20. Further calculations

Video (Lemma 20.1 and Proposition 20.2)

Suppose that n2. We previously noted that n{0} is homotopy equivalent to Sn-1 and so has H0Hn-1, with all other homology groups zero. One might ask what happens if we remove several points a1,,am from n instead of just removing the origin. One can answer this question by induction on m, but to make the induction work smoothly, it is convenient to generalise slightly. Instead of just considering n{a1,,am}, we will consider W{a1,,am}, where W is an arbitrary convex open subset of n. We first need the following lemma:

Lemma 20.1.

Suppose that X=UV, where U and V are open and connected, and UV is contractible. Then H0(X)= and Hk(X)=Hk(U)Hk(V) for all k1.


This follows from Proposition 15.2 (the truncated Mayer-Vietoris sequence). That result includes the fact that H0(X)=, and that there are exact sequences


for all k2, and also an exact sequence


As Hj(UV)=0 for all j>0, these exact sequences show that the map Hk(U)Hk(V)Hk(X) is an isomorphism for all k1. ∎

Proposition 20.2.

Let W be a convex open subset of n (with n2). Let A be a finite subset of W, with |A|=m say, and put X=WA. Then H0(X)= and Hn-1(X)=m and Hk(X)=0 for k{0,n-1}.


If m=0 then X=W so X is contractible so the claim is clear. Suppose instead that m=1 so A={a} say. Define r:XSn-1 by r(x)=(x-a)/x-a. Next, as W is open we can choose ϵ>0 such that the open ball of radius ϵ around a is contained in W. We can then define j:Sn-1X by j(y)=a+ϵy/2. We find that rj is the identity, and that jr is homotopic to the identity by a straight-line homotopy, so X is homotopy equivalent to Sn-1; the claim follows from this.

We now suppose that m>1 and that we have already proved the claim for sets of size less than m. We have only a finite number of vectors a-a with a,aA and aa; choose any unit vector u that is not perpendicular to any of these. We then find that the dot products u.a (for aA) are all different, so we can list the elements of A as a1,,am with u.a1<< Choose constants p,q with<p<q< Put V={xX|u.x<q} and U={xX|u.x>p}, so U and V are open with UV=X. The space V is obtained by removing {a1,,am-1} from the convex set {wW|u.w<q}, so the homology of V is given by the induction hypothesis. The space U is obtained by removing am from the set {wW|u.w>p}, so the homology of U is given by the case m=1. The intersection UV is just the convex set {wW|p<u.w<q} (with no points removed), so it is contractible. Lemma 20.1 therefore gives H0(X)= and Hk(X)=Hk(U)Hk(V) for all k>0, and the induction step is clear from this. ∎

Video (Examples 20.3 and 20.4)

Example 20.3.

Let Ln be the union of n adjacent squares arranged horizontally. The case n=3 is illustrated below.

Let U be the space obtained by removing the rightmost vertical edge; this is easily seen to be homotopy equivalent to Ln-1. Let V be the space obtained by removing Ln-2, leaving just the rightmost square with two extra edges attached; this is easily seen to be homotopy equivalent to S1. The intersection UV is a sideways H shape, and is easily seen to be contractible. We can therefore apply Lemma 20.1 to see that Hk(Ln)=Hk(Ln-1)Hk(S1) for all k>0. It follows inductively that H0(Ln)= and H1(Ln)=n and Hk(Ln)=0 for k>1. Alternatively, we can let ap be the centre of the p’th square, so Ln2{a1,,an}. It can be shown that the inclusion i:Ln2{a1,,an} is a homotopy equivalence. The homology of Ln can therefore be obtained from Proposition 20.2; it is easy to see that this gives the same answer.

Example 20.4.

Let Wn consist of n circles joined together at a single point. We can write this as UV as illustrated below. Given this, essentially the same argument as in Example 20.3 gives H0(Wn)= and H1(Wn)=n and Hk(Wn)=0 for k2.

Video (Proposition 20.5 and Remark 20.6)

Proposition 20.5.

The homology of the torus T=S1×S1 is given by H0(T)=H2(T)= and H1(T)=2 and Hk(T)=0 for k3.


We can form the torus by taking a square, gluing the top to the bottom and the left edge to the right edge. This is illustrated by the left hand picture below. The other pictures show two open sets U,VT with UV=T.

The set V is contractible, so H0(V)= and Hk(V)=0 for k>0. The set UV is homotopy equivalent to S1, so H0(UV)= and H1(UV)=.[u] and Hk(UV)=0 for k>1. It might appear that U is also homotopy equivalent to S1, but that is misleading because the edges are glued together. The linked demonstration makes it clear that U is the union of two circular bands, whose intersection is a filled square (and so is contractible). It therefore follows from Lemma 20.1 that H0(U)= and H1(U)= and Hk(U)=0 for k>1. More specifically, the two bands are thickenings of the loops v and w, so H1(U)={[v],[w]}. All the relevant spaces are connected, so H0(T)= and we have a truncated Mayer-Vietoris sequence as in Proposition 15.2. After filling in the known groups, we see that the tail end of this sequence is as follows:


In U, we can deform u outwards to the edge of the square (which does not change the homology class). It then becomes equal to the join v*w*v¯*w¯. As discussed in Proposition 10.29, this is homologous to v+w-v-w=0. This proves that i*=0, so ker(i*)=.[u] and img(i*)=0. As the sequence is exact, it follows that img(δ)=.[u] and ker(k*)=0, so the maps δ and k* are isomorphisms, so H1(T)2 and H2(T). For k3 we have an exact sequence


which shows that Hk(T)=0. ∎

Remark 20.6.

As an alternative, we could write the torus as the union of the following open sets:

Here U and V are both homotopy equivalent to S1, whereas UV is homotopy equivalent to the union of two disjoint copies of S1. We leave it to the reader to understand how the resulting Mayer-Vietoris sequence gives the same answer as before. This method can be generalised to calculate the homology of the d-dimensional torus Td=(S1)d=S1××S1. The answer is that Hk(Td)(dk) for all k, but we will not give the details here.

Proposition 20.7.

For the real projective plane P2 we have H0(P2)= and H1(P2)=/2 and Hk(P2)=0 for k2.


Recall (perhaps from the Knots and Surfaces course) that we can form P2 by taking a disc as shown on the left below, and identifying opposite points on the boundary circle. In particular, the two points marked a are the same, and each point on the upper red semicircle is identified with the corresponding point on the lower blue semicircle, so the two paths marked v are the same. We consider P2 as the union of the indicated open sets U and V.

It is clear that V is contractible, so H0(V)= and Hk(V)=0 for k1. It is also clear that that UV is homotopy equivalent to a circle, so H0(UV)= and H1(UV)=.[u] and Hk(UV)=0 for k2.

Next, we claim that U is homeomorphic to a Möbius strip, and therefore homotopy equivalent to a circle. One way to see this is to use the following animated diagram:

Alternatively, we can argue using the pictures below. The first one shows U together with two extra edges marked p and q. If we cut along these, we get the middle picture. If we flip the top half over and glue it to the bottom half along v, we get the third picture, which is the standard gluing diagram for a Möbius strip.

Either way, it follows that H0(U)= and H1(U)=.[v] and Hk(U)=0 for k2. We next need to understand the map i*:H1(UV)H1(U). In U, we can push the loop u out to the boundary by a loop homotopy, which does not change the homology class. The deformed loop then covers both copies of v, so we have i*([u])=2[v]. After taking account of the known homology groups, the end of the truncated Mayer-Vietoris sequence looks like this:

As i*([u])=2[v] we see that ker(i*)=0 and so exactness forces H2(P2) to be zero. Also, the map k* is surjective with kernel given by the image of i*, which is 2.[v], so H1(P2)/2. For k3 we also have exact sequences


so Hk(P2)=0. As P2 is connected, we have H0(P2)=. ∎

Example 20.8.

Now consider a closed surface X presented as in the Knots and Surfaces course, by gluing edges of a polygon according to a surface word. We illustrate this using the standard word aba-1b-1cdc-1d-1 for an orientable surface of genus 2, but the method works for any word satisfying the usual conditions of surface theory. We can calculate H*(X) using essentially the same method that we used for the torus. The conclusion is that H0(X)=H2(X)=, and Hk(X)=0 for k3, and H1(X)=2g, where g is the genus. In particular, in the illustrated case we have H1(X)=4. To see this, we use open sets U and V illustrated below.

As with the case of the torus, the space V is contractible and the space UV is homotopy equivalent to a circle, so their homology is easy to understand. The space U apparently has 8 edges and 8 marked vertices. However, the edges are glued together in pairs, in such a way that all vertices get glued together; the result is just the space W4 from Example 20.4. The full space U consists of W4 with a fringe attached, but that does not affect the homotopy type, so we have H*(U)=H*(W4). This means that H0(U)= and H1(U)=4 and Hk(U)=0 for k2. If we let u be a loop once around UV, then in U we see that u becomes a*b*a¯*b¯*c*d*c¯*d¯. This has homology class [a]+[b]-[a]-[b]+[c]+[d]-[c]-[d]=0, so the homomorphism i*:H1(UV)H1(U) is zero. We now have an exact sequence


and the claimed description of H*(X) follows easily from this.

Proposition 20.9.

The group H1(P1) is isomorphic to , but for n2 the group H1(Pn) is isomorphic to /2. Moreover, for k>n we have Hk(Pn)=0.


We saw in Example 7.23 that P1 is homeomorphic to S1, so H1(P1). The case n=2 is given by Proposition 20.7. We have an inclusion i:SnSn+1 given by i(x0,,xn)=(x0,,xn,0). This satisfies i(-x)=-i(x), so it induces an inclusion Sn/{±1}Sn+1/{±1}, or in other words PnPn+1; we again call this i. It will be enough to check that the map i*:H1(Pn)H1(Pn+1) is an isomorphism for n2.

Recall that Sn+1={xn+2|x2=1}. We can identify n+2 with n+1×; then


The subspace where z=0 can be identified with Sn. Now let U+ and U- be the subspaces where z>0 and z<0 respectively, and put U~=U+U-. We also put


Now let π:Sn+1Sn+1/{±1}=Pn+1 be the quotient projection (so π(x)=π(x) iff x=±x). Put U=π(U~) and V=π(V~). These sets are easily seen to be open with respect to the quotient topology on Pn+1.

Let U be the open ball of radius one in n+1. We have maps


where f(y)=(y,1-y2), and one can check that both of these are homeomorphisms. It follows that U is contractible. The same maps also give homeomorphisms


It follows that UV is homotopy equivalent to Sn.

Next, let j:SnV~ be the inclusion, and define r:V~Sn by r(y,z)=y/y (which is valid because y0 when (y,z)V~). This satisfies rj=id:SnSn. We also define h:[0,1]×V~V~ by


this gives a homotopy between jr and the identity. These maps satisfy j(-x)=-j(x) and r(-x)=-r(x) and h(t,-x)=-h(t,x) so they induce maps j:PnV and r:VPn and h:[0,1]×VV. Using this we see that j is a homotopy equivalence, so H1(V)=H1(Pn).

The spaces U, V, UV and UV=Pn+1 are all path connected, so we have a truncated Mayer-Vietoris sequence


Here UV is homotopy equivalent to Sn with n2 so H1(UV)=0. The space U is contractible so H1(U)=0. The space V is homotopy equivalent to Pn, so we can assume inductively that H1(V)=/2. Exactness of the sequence now implies that H1(Pn+1)=/2 as required.

Now suppose that k>n+1. We have a Mayer-Vietoris sequence


or equivalently


As k>n+1 we have Hk-1(Sn)=0. Also, we can assume inductively that Hj(Pn)=0 for all j>n, so in particular Hk(Pn)=0. It follows by exactness that Hk(Pn+1)=0 as claimed. ∎

Remark 20.10.

The full story is as follows. For m0 we have

Hk(P2m+1) ={ if k=0 or k=2m+1/2 if 0<k<2m and k is odd 0 otherwise.
Hk(P2m+2) ={ if k=0/2 if 0<k<2m+2 and k is odd 0 otherwise.

Equivalently, let P(n) be the following chain complex:


There are copies of in degrees 0 to n inclusive, and the differentials alternate between zero and multiplication by 2. Then H*(Pn)H*(P(n)) for all n. This can be proved using the same Mayer-Vietoris sequence as mentioned above, but some extra work is needed to determine the maps in that sequence.