MAS61015 Algebraic Topology

11. Homology of the punctured plane

Later we will prove that for all n2 we have

Hk(n{0})=Hk(Sn-1)={ if k=0 or k=n-10 otherwise. 

For this we will need the Mayer-Vietoris sequence, which is a very important and useful tool, but it will take some work to set that up. In this section, we outline a different approach which is more direct and elementary but which works only for n=2. We will identify 2 with and write ×={0}, so our main task will be to prove that H1(×)=.

Video (Definition 11.2 to Theorem 11.11)

Definition 11.1.

Let z× be a nonzero complex number. This can be expressed as z=reiθ for a unique pair or real numbers r,θ with r>0 and -π<θπ. We put plog(z)=log(r)+iθ, and call this the principal logarithm of z.

Note that plog:× and exp(plog(z))=z for all z, but plog is not continuous (because for small ϵ>0 we have plog(-1+iϵ)iπ but plog(-1-iϵ)-iπ). This cannot be fixed by adjusting the definitions: there is no continuous map f:× with exp(f(z))=z for all z. To work around this we make the following definition:

Definition 11.2.

For any z× we put


Any element of LOG(z) will be called a logarithm of z. More generally, suppose we have a topological space T and a continuous map u:T×. By a continuous logarithm of u we mean a continuous map u~:T with expu~=u, or equivalently u~(t)LOG(u(t)) for all t.

Note that the cosets iπ+2πi and -iπ+2πi are the same, and that LOG(-1+iϵ) is close to this coset for all small ϵ, independent of whether ϵ is positive or negative. Thus, LOG(z) depends continuously on z even though plog(z) does not.

Given a continuous map u:T×, we could attempt to define a continuous logarithm of u by u~=plogu. This works provided that the image u(T) does not touch the negative real axis where plog is discontinuous. If u(T) does touch the negative real axis then it may be possible to find a continuous logarithm by a different method, but in some cases, no continuous logarithm exists.

Lemma 11.3.

Let u:[0,1]× be continuous, and suppose that xLOG(u(0)). Then there is a unique continuous logarithm u~:[0,1] with u~(0)=x.

We will prove this properly later when we come to discuss covering maps.

Sketch proof.

If we choose N sufficently large, then when |s-t|1/N the points u(s)/u(t) will be close to 1 in × and so will be far from the negative real axis where plog is discontinuous. We can thus define


This is a continuous function of t. When t=0 we see that all the terms in the sum are plog(u(0)/u(0))=plog(1)=0, so u~(0)=x. In general we have


and most of the terms in the product cancel out leaving only exp(u~(t))=u(t). ∎

Corollary 11.4.

Let KN be convex, and suppose that kK. Let u:K× be continuous, and suppose that xLOG(u(k)). Then there is a unique continuous logarithm u~:K with u~(k)=x. (In particular, this applies when K=Δd for some d.)

Sketch proof.

For mK we can define vm:[0,1]× by vm(t)=u(tm+(1-t)k). By the lemma, there is a unique continuous logarithm v~m:[0,1] with v~m(0)=x. We define u~(m)=v~m(1), so exp(u~(m))=exp(v~m(1))=vm(1)=u(m). We have vk(t)=u(k) for all t so v~k must be the constant path at x so u~(k)=v~k(1)=x. With some work one can check that u~ is continuous. ∎

Definition 11.5.

Given any path u:Δ1× we define


where u~ is any continuous logarithm of u. (This is well-defined, because any two continuous logarithms differ by a constant of the form 2nπi, and the constant cancels out when we calculate u~(e1)-u~(e0).) We then extend this linearly to get a homomorphism ω:C1(×), given by

Example 11.6.

The standard loop un:Δ1× of winding number n is given by un(1-t,t)=exp(2πint). The obvious continuous logarithm is u~n(t)=2πint, and using this we get ω(un)=n.

Lemma 11.7.

For u:Δ2× we have ω((u))=0. Thus, we have ω(B1(×))=0.


For i=0,1,2 we put vi=uδi:Δ1×, so (u)=v0-v1+v2. By Corollary 11.4, we can choose a continuous logarithm u~:Δ2 for u. We then note that the map v~i=u~δi:Δ1 is a continuous logarithm for for vi, so ω(vi)=(v~i(e1)-v~i(e0))/(2πi). This gives


However, we have

δ0(e1) =e2 δ1(e1) =e2 δ2(e1) =e1
δ0(e0) =e1 δ1(e0) =e0 δ2(e0) =e0,

so the above expression becomes


so ω((u))=0 as claimed. More generally, if uC2(×) then u=n1u1++nrur for some integers ni and maps ui:Δ2×, and this gives


as before. Thus, if wB1(×) then w=(u) for some uC2(×) giving ω(w)=ω((u))=0 as claimed. ∎

Definition 11.8.

We now define homomorphisms β:× and γ:C0(×)× by β(z)=exp(2πiz) and


(Here we regard and C0(×) as groups under addition and × as a group under multiplication, so to say that β and γ are homomorphisms means that β(w+z)=β(w)β(z) and γ(u+v)=γ(u)γ(v); it is easy to see that both of these identities are valid.)

Lemma 11.9.

The following square commutes (or in other words, β(ω(u))=γ((u)) for all uC1(×)).


In general u will be a -linear combination of paths in ×, but all the maps are homomorphisms, so it will be enough to consider the case where u:Δ1× is just a single path. We then have (u)=u(e1)-u(e0)C0(×) and so γ((u))=u(e1)/u(e0)×. Now choose a continuous logarithm u~:Δ1 for u. By definition we have ω(u)=(u~(e1)-u~(e0))/(2πi), so


Corollary 11.10.

We have ω(Z1(×))= and ω(B1(×))=0, so ω induces a homomorphism ω¯:H1(×) given by ω¯(z+B1(×))=ω(z).


Suppose that zZ1(×), so (z)=0, so γ((z))=γ(0)=1. By the Lemma we then have β(ω(z))=1, or in other words exp(2πiω(z))=1, so ω(z). As in Example 11.6 we also have standard loops unZ1(×) with ω(un)=n, so the image ω(Z1(×)) is the whole group . We saw in Lemma 11.7 that ω(B1(×))=0, and it follows that the rule ω¯(z+B1(×))=ω(z) gives a well-defined homomorphism from the quotient group H1(×)=Z1(×)/B1(×) to . ∎

Theorem 11.11.

The homomorphism ω¯:H1(×) is an isomorphism.


We have already remarked that ω¯(un+B1(×))=ω(un)=n for all n; this shows that ω¯ is surjective. Now suppose we have hH1(×) with ω(h)¯=0. By Proposition 10.29, we can find a loop u:Δ1× based at 1× with h=[u]. By Lemma 11.3, there is a unique continuous logarithm u~:Δ1 with u~(e0)=0. We then have ω(u)=(u~(e1)-u~(e0))/(2πi)=u~(e1)/(2πi). However, we also know that ω(u)=ω¯([u])=ω¯(h)=0, so we must have u~(e1)=0 as well. We now define v~:Δ2 by

v~(t0,t1,t2)={(1-t0)u~(t1/(1-t0),t2/(1-t0)) if t0<10 if t0=1.

We leave it to the reader to check that v~ is continuous even at e0. (A full proof of a more general fact will be given later.) We then define v=expv~:Δ2×. It is easy to see that v~ is a filling in for u~, and thus that v is a filling in for u, so [u]=0 in H1(X) by Lemma 10.31, or in other words h=0. This proves that ω¯ is also injective, and so is an isomorphism as claimed. ∎