MAS61015 Algebraic Topology 19 Construction of the Mayer-Vietoris sequence 21 The Jordan Curve Theorem

20. Further calculations

Video (Lemma 20.1 and Proposition 20.2)

Suppose that $n\geq 2$ . We previously noted that ${\mathbb{R}}^{n}\setminus\{0\}$ is homotopy equivalent to $S^{n-1}$ and so has $H_{0}\simeq H_{n-1}\simeq{\mathbb{Z}}$ , with all other homology groups zero. One might ask what happens if we remove several points $a_{1},\dotsc,a_{m}$ from ${\mathbb{R}}^{n}$ 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 ${\mathbb{R}}^{n}\setminus\{a_{1},\dotsc,a_{m}\}$ , we will consider $W\setminus\{a_{1},\dotsc,a_{m}\}$ , where $W$ is an arbitrary convex open subset of ${\mathbb{R}}^{n}$ . We first need the following lemma:

Lemma 20.1.

Suppose that $X=U\cup V$ , where $U$ and $V$ are open and connected, and $U\cap V$ is contractible. Then $H_{0}(X)={\mathbb{Z}}$ and $H_{k}(X)=H_{k}(U)\oplus H_{k}(V)$ for all $k\geq 1$ .

Proof.

This follows from Proposition 15.2 (the truncated Mayer-Vietoris sequence). That result includes the fact that $H_{0}(X)={\mathbb{Z}}$ , and that there are exact sequences

H_{k}(U\cap V)\to H_{k}(U)\oplus H_{k}(V)\to H_{k}(X)\to H_{k-1}(U\cap V)

for all $k\geq 2$ , and also an exact sequence

H_{1}(U\cap V)\to H_{1}(U)\oplus H_{1}(V)\to H_{1}(X)\to 0.

As $H_{j}(U\cap V)=0$ for all $j>0$ , these exact sequences show that the map $H_{k}(U)\oplus H_{k}(V)\to H_{k}(X)$ is an isomorphism for all $k\geq 1$ . ∎

Proposition 20.2.

Let $W$ be a convex open subset of ${\mathbb{R}}^{n}$ (with $n\geq 2$ ). Let $A$ be a finite subset of $W$ , with $|A|=m$ say, and put $X=W\setminus A$ . Then $H_{0}(X)={\mathbb{Z}}$ and $H_{n-1}(X)={\mathbb{Z}}^{m}$ and $H_{k}(X)=0$ for $k\not\in\{0,n-1\}$ .

Proof.

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 $r\colon X\to S^{n-1}$ by $r(x)=(x-a)/\|x-a\|$ . Next, as $W$ is open we can choose $\epsilon>0$ such that the open ball of radius $\epsilon$ around $a$ is contained in $W$ . We can then define $j\colon S^{n-1}\to X$ by $j(y)=a+\epsilon y/2$ . We find that $r j$ is the identity, and that $j r$ is homotopic to the identity by a straight-line homotopy, so $X$ is homotopy equivalent to $S^{n-1}$ ; 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 $a-a^{\prime}$ with $a,a^{\prime}\in A$ and $a\neq a^{\prime}$ ; choose any unit vector $u$ that is not perpendicular to any of these. We then find that the dot products $u . a$ (for $a\in A$ ) are all different, so we can list the elements of $A$ as $a_{1},\dotsc,a_{m}$ with $u.a_{1}<\dotsb<u.a_{m}$ . Choose constants $p, q$ with $u.a_{m-1}<p<q<u.a_{m}$ . Put $V=\{x\in X\;|\;u.x<q\}$ and $U=\{x\in X\;|\;u.x>p\}$ , so $U$ and $V$ are open with $U\cup V=X$ . The space $V$ is obtained by removing $\{a_{1},\dotsc,a_{m-1}\}$ from the convex set $\{w\in W\;|\;u.w<q\}$ , so the homology of $V$ is given by the induction hypothesis. The space $U$ is obtained by removing $a_{m}$ from the set $\{w\in W\;|\;u.w>p\}$ , so the homology of $U$ is given by the case $m=1$ . The intersection $U\cap V$ is just the convex set $\{w\in W\;|\;p<u.w<q\}$ (with no points removed), so it is contractible. Lemma 20.1 therefore gives $H_{0}(X)={\mathbb{Z}}$ and $H_{k}(X)=H_{k}(U)\oplus H_{k}(V)$ for all $k>0$ , and the induction step is clear from this. ∎

Video (Examples 20.3 and 20.4)

Example 20.3.

Let $L_{n}$ 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 $L_{n-1}$ . Let $V$ be the space obtained by removing $L_{n-2}$ , leaving just the rightmost square with two extra edges attached; this is easily seen to be homotopy equivalent to $S^{1}$ . The intersection $U\cap V$ is a sideways H shape, and is easily seen to be contractible. We can therefore apply Lemma 20.1 to see that $H_{k}(L_{n})=H_{k}(L_{n-1})\oplus H_{k}(S^{1})$ for all $k>0$ . It follows inductively that $H_{0}(L_{n})={\mathbb{Z}}$ and $H_{1}(L_{n})={\mathbb{Z}}^{n}$ and $H_{k}(L_{n})=0$ for $k>1$ . Alternatively, we can let $a_{p}$ be the centre of the $p$ ’th square, so $L_{n}\subseteq{\mathbb{R}}^{2}\setminus\{a_{1},\dotsc,a_{n}\}$ . It can be shown that the inclusion $i\colon L_{n}\to{\mathbb{R}}^{2}\setminus\{a_{1},\dotsc,a_{n}\}$ is a homotopy equivalence. The homology of $L_{n}$ can therefore be obtained from Proposition 20.2; it is easy to see that this gives the same answer.

Example 20.4.

Let $W_{n}$ consist of $n$ circles joined together at a single point. We can write this as $U\cup V$ as illustrated below. Given this, essentially the same argument as in Example 20.3 gives $H_{0}(W_{n})={\mathbb{Z}}$ and $H_{1}(W_{n})={\mathbb{Z}}^{n}$ and $H_{k}(W_{n})=0$ for $k\geq 2$ .

Video (Proposition 20.5 and Remark 20.6)

Proposition 20.5.

The homology of the torus $T=S^{1}\times S^{1}$ is given by $H_{0}(T)=H_{2}(T)={\mathbb{Z}}$ and $H_{1}(T)={\mathbb{Z}}^{2}$ and $H_{k}(T)=0$ for $k\geq 3$ .

Proof.

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 $U,V\subseteq T$ with $U\cup V=T$ .

Interactive demo

The set $V$ is contractible, so $H_{0}(V)={\mathbb{Z}}$ and $H_{k}(V)=0$ for $k>0$ . The set $U\cap V$ is homotopy equivalent to $S^{1}$ , so $H_{0}(U\cap V)={\mathbb{Z}}$ and $H_{1}(U\cap V)={\mathbb{Z}}.[u]$ and $H_{k}(U\cap V)=0$ for $k>1$ . It might appear that $U$ is also homotopy equivalent to $S^{1}$ , 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 $H_{0}(U)={\mathbb{Z}}$ and $H_{1}(U)={\mathbb{Z}}\oplus{\mathbb{Z}}$ and $H_{k}(U)=0$ for $k>1$ . More specifically, the two bands are thickenings of the loops $v$ and $w$ , so $H_{1}(U)={\mathbb{Z}}\{[v],[w]\}$ . All the relevant spaces are connected, so $H_{0}(T)={\mathbb{Z}}$ 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:

0\to H_{2}(T)\xrightarrow{\delta}{\mathbb{Z}}.[u]\xrightarrow{i_{*}}{\mathbb{Z% }}\{[v],[w]\}\xrightarrow{k_{*}}H_{1}(T)\to 0.

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 $v*w*\overline{v}*\overline{w}$ . As discussed in Proposition 10.29, this is homologous to $v+w-v-w=0$ . This proves that $i_{*}=0$ , so $\ker(i_{*})={\mathbb{Z}}.[u]$ and $\operatorname{img}(i_{*})=0$ . As the sequence is exact, it follows that $\operatorname{img}(\delta)={\mathbb{Z}}.[u]$ and $\ker(k_{*})=0$ , so the maps $\delta$ and $k_{*}$ are isomorphisms, so $H_{1}(T)\simeq{\mathbb{Z}}^{2}$ and $H_{2}(T)\simeq{\mathbb{Z}}$ . For $k\geq 3$ we have an exact sequence

0=H_{k}(U)\oplus H_{k}(V)\to H_{k}(X)\xrightarrow{\delta}H_{k-1}(U\cap V)=0,

which shows that $H_{k}(T)=0$ . ∎

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 $S^{1}$ , whereas $U\cap V$ is homotopy equivalent to the union of two disjoint copies of $S^{1}$ . 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 $T_{d}=(S^{1})^{d}=S^{1}\times\dotsb\times S^{1}$ . The answer is that $H_{k}(T_{d})\simeq{\mathbb{Z}}^{\binom{d}{k}}$ for all $k$ , but we will not give the details here.

Proposition 20.7.

For the real projective plane ${\mathbb{R}}P^{2}$ we have $H_{0}({\mathbb{R}}P^{2})={\mathbb{Z}}$ and $H_{1}({\mathbb{R}}P^{2})={\mathbb{Z}}/2$ and $H_{k}({\mathbb{R}}P^{2})=0$ for $k\geq 2$ .

Proof.

Video

Recall (perhaps from the Knots and Surfaces course) that we can form ${\mathbb{R}}P^{2}$ 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 ${\mathbb{R}}P^{2}$ as the union of the indicated open sets $U$ and $V$ .

It is clear that $V$ is contractible, so $H_{0}(V)={\mathbb{Z}}$ and $H_{k}(V)=0$ for $k\geq 1$ . It is also clear that that $U\cap V$ is homotopy equivalent to a circle, so $H_{0}(U\cap V)={\mathbb{Z}}$ and $H_{1}(U\cap V)={\mathbb{Z}}.[u]$ and $H_{k}(U\cap V)=0$ for $k\geq 2$ .

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:

Interactive demo

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 $H_{0}(U)={\mathbb{Z}}$ and $H_{1}(U)={\mathbb{Z}}.[v]$ and $H_{k}(U)=0$ for $k\geq 2$ . We next need to understand the map $i_{*}\colon H_{1}(U\cap V)\to H_{1}(U)$ . 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 $H_{2}({\mathbb{R}}P^{2})$ to be zero. Also, the map $k_{*}$ is surjective with kernel given by the image of $i_{*}$ , which is $2{\mathbb{Z}}.[v]$ , so $H_{1}({\mathbb{R}}P^{2})\simeq{\mathbb{Z}}/2$ . For $k\geq 3$ we also have exact sequences

0=H_{k}(U)\oplus H_{k}(V)\to H_{k}({\mathbb{R}}P^{2})\to H_{k-1}(U\cap V)=0,

so $H_{k}({\mathbb{R}}P^{2})=0$ . As ${\mathbb{R}}P^{2}$ is connected, we have $H_{0}({\mathbb{R}}P^{2})={\mathbb{Z}}$ . ∎

Example 20.8.

Video

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 $aba^{-1}b^{-1}cdc^{-1}d^{-1}$ 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 $H_{0}(X)=H_{2}(X)={\mathbb{Z}}$ , and $H_{k}(X)=0$ for $k\geq 3$ , and $H_{1}(X)={\mathbb{Z}}^{2g}$ , where $g$ is the genus. In particular, in the illustrated case we have $H_{1}(X)={\mathbb{Z}}^{4}$ . 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 $U\cap V$ 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 $W_{4}$ from Example 20.4. The full space $U$ consists of $W_{4}$ with a fringe attached, but that does not affect the homotopy type, so we have $H_{*}(U)=H_{*}(W_{4})$ . This means that $H_{0}(U)={\mathbb{Z}}$ and $H_{1}(U)={\mathbb{Z}}^{4}$ and $H_{k}(U)=0$ for $k\geq 2$ . If we let $u$ be a loop once around $U\cap V$ , then in $U$ we see that $u$ becomes $a*b*\overline{a}*\overline{b}*c*d*\overline{c}*\overline{d}$ . This has homology class $[a]+[b]-[a]-[b]+[c]+[d]-[c]-[d]=0$ , so the homomorphism $i_{*}\colon H_{1}(U\cap V)\to H_{1}(U)$ is zero. We now have an exact sequence

0\to H_{2}(X)\xrightarrow{\delta}{\mathbb{Z}}\xrightarrow{i_{*}=0}{\mathbb{Z}}% ^{4}\xrightarrow{}H_{1}(X)\to 0,

and the claimed description of $H_{*}(X)$ follows easily from this.

Proposition 20.9.

The group $H_{1}({\mathbb{R}}P^{1})$ is isomorphic to ${\mathbb{Z}}$ , but for $n\geq 2$ the group $H_{1}({\mathbb{R}}P^{n})$ is isomorphic to ${\mathbb{Z}}/2$ . Moreover, for $k>n$ we have $H_{k}({\mathbb{R}}P^{n})=0$ .

Proof.

Video

We saw in Example 7.23 that ${\mathbb{R}}P^{1}$ is homeomorphic to $S^{1}$ , so $H_{1}({\mathbb{R}}P^{1})\simeq{\mathbb{Z}}$ . The case $n=2$ is given by Proposition 20.7. We have an inclusion $i\colon S^{n}\to S^{n+1}$ given by $i(x_{0},\dotsc,x_{n})=(x_{0},\dotsc,x_{n},0)$ . This satisfies $i(-x)=-i(x)$ , so it induces an inclusion $S^{n}/\{\pm 1\}\to S^{n+1}/\{\pm 1\}$ , or in other words ${\mathbb{R}}P^{n}\to{\mathbb{R}}P^{n+1}$ ; we again call this $i$ . It will be enough to check that the map $i_{*}\colon H_{1}({\mathbb{R}}P^{n})\to H_{1}({\mathbb{R}}P^{n+1})$ is an isomorphism for $n\geq 2$ .

Recall that $S^{n+1}=\{x\in{\mathbb{R}}^{n+2}\;|\;\|x\|^{2}=1\}$ . We can identify ${\mathbb{R}}^{n+2}$ with ${\mathbb{R}}^{n+1}\times{\mathbb{R}}$ ; then

S^{n+1}=\{(y,z)\in{\mathbb{R}}^{n+1}\times{\mathbb{R}}\;|\;\|y\|^{2}+z^{2}=1\}.

The subspace where $z=0$ can be identified with $S^{n}$ . Now let $U_{+}$ and $U_{-}$ be the subspaces where $z>0$ and $z<0$ respectively, and put $\widetilde{U}=U_{+}\cup U_{-}$ . We also put

\widetilde{V}=\{(y,z)\in S^{n+1}\;|\;y\neq 0\}=\{(y,z)\in S^{n+1}\;|\;-1<z<1\}% =S^{n+2}\setminus\{(0,1),(0,-1)\}.

Now let $\pi\colon S^{n+1}\to S^{n+1}/\{\pm 1\}={\mathbb{R}}P^{n+1}$ be the quotient projection (so $\pi(x)=\pi(x^{\prime})$ iff $x^{\prime}=\pm x$ ). Put $U=\pi(\widetilde{U})$ and $V=\pi(\widetilde{V})$ . These sets are easily seen to be open with respect to the quotient topology on ${\mathbb{R}}P^{n+1}$ .

Let $U^{\prime}$ be the open ball of radius one in ${\mathbb{R}}^{n+1}$ . We have maps

U^{\prime}\xrightarrow{f}U_{+}\xrightarrow{\pi}U,

where $f(y)=(y,\sqrt{1-\|y\|^{2}})$ , and one can check that both of these are homeomorphisms. It follows that $U$ is contractible. The same maps also give homeomorphisms

U^{\prime}\setminus\{0\}\xrightarrow{f}U_{+}\setminus\{(0,1)\}\xrightarrow{\pi% }U\cap V.

It follows that $U\cap V$ is homotopy equivalent to $S^{n}$ .

Next, let $j\colon S^{n}\to\widetilde{V}$ be the inclusion, and define $r\colon\widetilde{V}\to S^{n}$ by $r(y,z)=y/\|y\|$ (which is valid because $y\neq 0$ when $(y,z)\in\widetilde{V}$ ). This satisfies $r\circ j=\operatorname{id}\colon S^{n}\to S^{n}$ . We also define $h\colon[0,1]\times\widetilde{V}\to\widetilde{V}$ by

h(t,(y,z))=(y,tz)/\sqrt{\|y\|^{2}+t^{2}z^{2}};

this gives a homotopy between $j\circ r$ 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 $j\colon{\mathbb{R}}P^{n}\to V$ and $r\colon V\to{\mathbb{R}}P^{n}$ and $h\colon[0,1]\times V\to V$ . Using this we see that $j$ is a homotopy equivalence, so $H_{1}(V)=H_{1}({\mathbb{R}}P^{n})$ .

The spaces $U$ , $V$ , $U\cap V$ and $U\cup V={\mathbb{R}}P^{n+1}$ are all path connected, so we have a truncated Mayer-Vietoris sequence

H_{1}(U\cap V)\to H_{1}(U)\oplus H_{1}(V)\to H_{1}({\mathbb{R}}P^{n+1})\to 0.

Here $U\cap V$ is homotopy equivalent to $S^{n}$ with $n\geq 2$ so $H_{1}(U\cap V)=0$ . The space $U$ is contractible so $H_{1}(U)=0$ . The space $V$ is homotopy equivalent to ${\mathbb{R}}P^{n}$ , so we can assume inductively that $H_{1}(V)={\mathbb{Z}}/2$ . Exactness of the sequence now implies that $H_{1}({\mathbb{R}}P^{n+1})={\mathbb{Z}}/2$ as required.

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

H_{k}(U)\oplus H_{k}(V)\to H_{k}({\mathbb{R}}P^{n+1})\to H_{k-1}(U\cap V),

or equivalently

H_{k}({\mathbb{R}}P^{n})\to H_{k}({\mathbb{R}}P^{n+1})\to H_{k-1}(S^{n}).

As $k>n+1$ we have $H_{k-1}(S^{n})=0$ . Also, we can assume inductively that $H_{j}({\mathbb{R}}P^{n})=0$ for all $j>n$ , so in particular $H_{k}({\mathbb{R}}P^{n})=0$ . It follows by exactness that $H_{k}({\mathbb{R}}P^{n+1})=0$ as claimed. ∎

Remark 20.10.

The full story is as follows. For $m\geq 0$ we have

	$\displaystyle H_{k}({\mathbb{R}}P^{2m+1})$	$\displaystyle=\begin{cases}\hidden@noalign{}\textstyle{\mathbb{Z}}&\text{ if }% k=0\text{ or }k=2m+1\\ \hidden@noalign{}\textstyle{\mathbb{Z}}/2&\text{ if }0<k<2m\text{ and }k\text{% is odd }\\ \hidden@noalign{}\textstyle 0&\text{ otherwise. }\end{cases}$
	$\displaystyle H_{k}({\mathbb{R}}P^{2m+2})$	$\displaystyle=\begin{cases}\hidden@noalign{}\textstyle{\mathbb{Z}}&\text{ if }% k=0\\ \hidden@noalign{}\textstyle{\mathbb{Z}}/2&\text{ if }0<k<2m+2\text{ and }k% \text{ is odd }\\ \hidden@noalign{}\textstyle 0&\text{ otherwise. }\end{cases}$

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

\dotsb\xleftarrow{}0\xleftarrow{}0\xleftarrow{}{\mathbb{Z}}\xleftarrow{0}{% \mathbb{Z}}\xleftarrow{2}{\mathbb{Z}}\xleftarrow{0}{\mathbb{Z}}\xleftarrow{2}% \dotsb\xleftarrow{}{\mathbb{Z}}\xleftarrow{}0\xleftarrow{}0\xleftarrow{}\dotsb.

There are copies of ${\mathbb{Z}}$ in degrees $0$ to $n$ inclusive, and the differentials alternate between zero and multiplication by $2$ . Then $H_{*}({\mathbb{R}}P^{n})\simeq H_{*}(P(n))$ 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.