6. Seifert-van Kampen Theorem

Seifert-van Kampen Theorem

Theorem 6.1 (Seifert-van Kampen Theorem, Groupoid version). Let $X=U\cup V$ where $U,V\subset X$ are open. Then the following diagram is a pushout in the category $\mathsf{Groupoid}$.

Proof. Let $\mathcal{C}$ be a groupoid fitting into the diagram and we need to show that

Uniqueness:	Let $\gamma :I\to X$ be a path in $X$ with $x_t=\gamma (t)$. We subdivide $I$ (by its compactness) into $$0=t_0<t_1<\cdots <t_m=1$$such that ${\gamma }_i:=\gamma (t_{i-1},t_i)$ lies entirely in $U$ or $V$. Then $$F([\gamma ])=F([{\gamma }_m])\cdots F([{\gamma }_1])$$is determined uniquely in $\mathcal{C}$ as each term is.
Existence:	Given a path $\gamma$, we can define $F([\gamma ])$ using a subdivision of $I$ (or $\gamma$), where the result does not depend on the choice of the subdivision. We need to show that this is well-defined on homotopy class. This follows from a refined double subdivision of $I\times I$, as shown in the picture below.

Each square represents a homotopy lying entirely in either $U$ or $V$ and combining them together gives the require homotopy. $$F({\gamma }_1)\simeq F({\gamma }_1\star i_{x_0})\simeq F(i_{x_1}\star {\gamma }_2)\simeq F({\gamma }_2)$$$\square$

Theorem 6.2 (Seifert-van Kampen Theorem, Group version). Let $X=U\cup V$ where $U,V\subset X$ are open and $U$, $V$, $U\cap V$ are path connected. Let $x_0\in U\cap V$. Then the following diagram is a pushout in the category $\mathsf{Group}$.

Proof. Denote by $\underline{G}$ the groupoid with one object that comes from a group $G$.

For each $x\in X$, we fix a choice of $[{\gamma }_x]\in \mathrm{Hom}(x_0,x)$ such that ${\gamma }_x$ lies entirely in $U$ when $x\in U$ and ${\gamma }_x$ lies entirely in $V$ when $x\in V$. Note that this implies ${\gamma }_x$ lies entirely in $U\cap V$ when $x\in U\cap V$. Such choice can be achieved because $U$, $V$, $U\cap V$ are all path connected. Consider The following functors $$\begin{array}{rl}{\Pi }_1(U)& \to \underline{{\pi }_1(U,x_0)}\\ {\Pi }_1(V)& \to \underline{{\pi }_1(V,x_0)}\\ {\Pi }_1(U\cap V)& \to \underline{{\pi }_1(U\cap V,x_0)}\\ \gamma & \mapsto {\gamma }_{x_2}^{-1}\star \gamma \star {\gamma }_{x_1},\gamma \in \mathrm{Hom}(x_1,x_2)\end{array}$$These functors are all retracts in $\mathsf{Groupoid}$, in other words, the compositions $$\begin{array}{rl}\underline{{\pi }_1(U,x_0)}& \to {\Pi }_1(U)\to \underline{{\pi }_1(U,x_0)}\\ \underline{{\pi }_1(V,x_0)}& \to {\Pi }_1(V)\to \underline{{\pi }_1(V,x_0)}\\ \underline{{\pi }_1(U\cap V,x_0)}& \to {\Pi }_1(U\cap V)\to \underline{{\pi }_1(U\cap V,x_0)}\end{array}$$are all identity functors.

Suppose there is a group $G$ that fits into the following commutative diagram: By Theorem 6.1, we have the following morphism $F$ Thus we obtain a morphism $$\underline{{\pi }_1(X,x_0)}\hookrightarrow {\Pi }_1(X)\stackrel{F}{\to }\underline{G}$$which fits into a commutative diagram Since $\mathsf{Group}$ is a full subcategory of $\mathsf{Groupoid}$¹, the theorem follows.

$\square$

We also have a relative version.

Definition 6.3. Let $A\subset X$, we define ${\Pi }_1(X,A)$ be the full subcategory of ${\Pi }_1(X)$ consisting of objects in $A$.

For instance, when $A=\{x_0\}$, we have$${\Pi }_1(X,x_0)={\pi }_1(X,x_0).$$

Theorem 6.4. Let $X=U\cup V$, $U$, $V$ be open and $A\subset X$ intersects each path connected components of $U$, $V$, $U\cap V$. Then we have a pushout

Example 6.5. For the Figure-8 in Example 3.9, which is $S^1\vee S^1$. It can be decomposed into $U$, $V$ as follows Since $U$, $V$ are homotopic to $S^1$, and $U\cap V$ is homotopic to a point, Seifert-van Kampen Theorem implies $${\pi }_1(S^1\vee S^1)={\pi }_1(S^1)\star {\pi }_1(S^1)=\mathbb{Z}\star \mathbb{Z}.$$In general, we have $${\pi }_1\left(\underset{i=1}{\overset{n}{\bigvee }}{S}^1\right)=\underset{n}{\underbrace{\mathbb{Z}\star \cdots \star \mathbb{Z}}}.$$

Example 6.6. Consider the $2$-sphere $S^2=D_1\cup D_2$ where $D_1\cong D_2$ are open disks and $D_0=D_1\cap D_2$ is an annulus. Here $D_i$ is an open neighbourhood of $X_i$ for $i=0,1,2$. Since ${\pi }_1(D_1)={\pi }_1(D_2)$, ${\pi }_1(D_0)={\pi }_1(S^1)=\mathbb{Z}$, we deduce that $${\pi }_1(S^2)=(1\star 1)/\mathbb{Z}=1.$$A similar argument shows that $${\pi }_1(S^n)=1,n\ge 2.$$

Example 6.7. Let us identify $X=S^1$ with the unit circle in ${\mathbb{R}}^2$. Consider $$U=\{(x,y)\in S^1|y>-1/2\},V=\{(x,y)\in S^1|y<1/2\}$$and $A=\{(\pm 1,0)\}$. Then we obtain a pushout by Theorem 6.4 This implies that the groupoid ${\Pi }_1(S^1,A)$ contains two objects $a_1=(1,0)$, $a_2=(-1,0)$ with morphisms $$\begin{array}{rl}{\mathrm{Hom}}_{{\Pi }_1(S^1,A)}(a_1,a_1)& =\{({\gamma }_{-}{\gamma }_{+}{)}^n{\}}_{n\in \mathbb{Z}}\\ {\mathrm{Hom}}_{{\Pi }_1(S^1,A)}(a_1,a_2)& =\{({\gamma }_{+}{\gamma }_{-}{)}^n{\gamma }_{+}{\}}_{n\in \mathbb{Z}}\\ {\mathrm{Hom}}_{{\Pi }_1(S^1,A)}(a_2,a_1)& =\{({\gamma }_{-}{\gamma }_{+}{)}^n{\gamma }_{-}{\}}_{n\in \mathbb{Z}}\\ {\mathrm{Hom}}_{{\Pi }_1(S^1,A)}(a_2,a_2)& =\{({\gamma }_{+}{\gamma }_{-}{)}^n{\}}_{n\in \mathbb{Z}}.\end{array}$$Here ${\gamma }_{+}$ represents the semi-circle from $(1,0)$ to $(-1,0)$ anti-clockwise, and ${\gamma }_{-}$ represents the semi-circle from $(-1,0)$ to $(1,0)$ anti-clockwise.

Example 6.8. Consider the closed orientable surface ${\Sigma }_g$ of genus $g$, which admits a polygon presentation $$P=a_1{b}_1{a}_1^{-1}{b}_1^{-1}\cdots a_g{b}_g{a}_g^{-1}{b}_g^{-1}.$$Here is a figure for $g=2$. The edges of the polygon form $V_{2g}={\bigvee }_{i=1}^{2g}{S}^1$. Let $U$ be the interior of the polygon and $V$ be a small open neighbourhood of $V_{2g}$. Then $U\cap V$ is an annulus, which is homotopic to $S^1$ with the generator $P$ as above. Thus$${\pi }_1({\Sigma }_g)=\left(\coprod _{i=1}^{2g}\mathbb{Z}\right)\star 0/\mathbb{Z}=⟨a_i,b_i|i=1,\cdots ,g⟩/\left(a_1{b}_1{a}_1^{-1}{b}_1^{-1}\cdots a_g{b}_g{a}_g^{-1}{b}_g^{-1}\right).$$

Example 6.9. Using the polygon presentation $P=a^2$, we can similarly compute ${\pi }_1(\mathbb{R}{\mathbb{P}}^2)=\mathbb{Z}/2\mathbb{Z}$.

The Jordan Curve Theorem

We give an application of Seifert-van Kampen Theorem to prove the Jordan Curve Theorem. This is an example which sounds totally obvious intuitively, but turns out to be very difficult to prove rigorously.

Definition 6.10. A simple closed curve is a subset of ${\mathbb{R}}^2$ (or $S^2$) which is homeomorphic to the circle $S^1$.

Theorem 6.11 (The Jordan Curve Theorem). Let $C$ be a simple closed curve in the sphere $S^2$. Then the complement of $C$ has exactly two connected compoments.

Proof. We sketch a proof here. Since $S^2$ is locally path connected, we would not distinguish connected and path connected here. By an arc, we mean a subset of $S^2$ which is homeomorphic to the interval $I$.

We first show that: $$\text{if }A\text{ is an arc in }{S}^2,\text{ then }{S}^2\setminus A\text{ is connected.}$$In fact, assume that there are two points $\{a,b\}$ which are disconnected in $S^2\setminus A$. Let us subdivide $A=A_1\cup A_2$ into two intervals where $A_1=[0,1/2]$, $A_2=[1/2,1]$ using the homeomorphism $A\cong [0,1]$. We argue that $a,b$ are disconnected in either $S^1\setminus A_1$ or $S^2\setminus A_2$. Let us choose a set $D$ which contains one point from each connected component of $S^2\setminus A$ and such that $\{a,b\}\subset D$. Apply Seifert-van Kampen Theorem to $V_1=S^2\setminus A_1$, $V_2=S^2\setminus A_2$, $V_1\cap V_2=S^2\setminus A$, we obtain a pushout in $\mathsf{Groupoid}$ Here $Y=V_1\cup V_2$ is the complement of a point in $S^2$. If $\{a,b\}$ is connected in both $V_1$ and $V_2$, then the pushout implies that there exists a nontrivial morphism via the composition $$a\stackrel{\text{in }{V}_1}{\to }b\stackrel{\text{in }{V}_1\cap V_2}{=}b\stackrel{\text{in }{V}_2}{\to }a.$$But this can not true since $Y$ is contractible. So let us assume $a,b$ are disconnected in $V_1=S^2\setminus A_1$. Run the above process replacing $A$ by $A_1$, and keep doing this, we end up with contradiction in the limit. This proves our claim above for the arc.

Secondly, we show that: $$\text{The complement of }C\text{ in }{S}^2\text{ is disconnected}.$$Otherwise, assume that $S^2\setminus C$ is connected. Let us divide $C=A_1\cup A_2$ into two intervals $A_1$, $A_2$ which intersect at two endpoints $\{a,b\}$. Let $U_1=S^2\setminus A_1$, $U_2=S^2\setminus A_2$, $U_1\cap U_2=S^2\setminus C$ and $X=U_1\cup U_2=S^2\setminus \{a,b\}$. Since $U_1$, $U_2$, $U_1\cap U_2$ are all connected, Seifert-van Kampen Theorem leads to a pushout in $\mathsf{Group}$ Observe ${\pi }_1(X)=\mathbb{Z}$. We show both ${\pi }_1(U_i)\to {\pi }_1(X)$ are trivial. This would lead to the contradiction.

Let us identify $S^2={\mathbb{R}}^2\cup \{\infty \}$ and assume $a=0$, $b=\infty$, so $A_1$ is parametrized by a path $\alpha$ from $0$ to $\infty$. Let $\gamma$ be an arbitrary loop in $U_1$, we need to show $\gamma$ becomes trivial in $X$. Let $R>0$ be sufficiently large such that $\gamma$ is contained in the ball of radius $R$ centered at the origin in ${\mathbb{R}}^2$. Consider the homotopy $$F(t,s)=\gamma (t)-\alpha (s),{\gamma }_s:=F(-,s).$$We have ${\gamma }_0=\gamma$. Assume that $|\alpha (t_0)|>R$, then ${\gamma }_{t_0}$ lies inside the ball of radius $R$ centered at $-\alpha (t_0)$, which is contractible in $X$. This implies that $\gamma$ is trivial in $X$. The same argument applies to $A_2$.

Finally, we show that: $$\text{The complement of }C\text{ in }{S}^2\text{ has exactly two connected components}.$$Let $C=A_1\cup A_2$ and $U_1$, $U_2$ as in the previous step. Let $D$ be a set which contains exactly one point from each connected component of $S^2\setminus C$. We have a pushout diagram in $\mathsf{Groupoid}$ Suppose that $D$ contains at least three points, say $c$, $d$, $e$. Since $U_1$, $U_2$ are connected, and points in $D$ are disconnected in $U_1\cap U_2$, the following two compositions $$c\stackrel{\text{in }{U}_1}{\to }d\stackrel{\text{in }{U}_1\cap U_2}{=}d\stackrel{\text{in }{U}_2}{\to }c\text{ and }c\stackrel{\text{in }{U}_1}{\to }e\stackrel{\text{in }{U}_1\cap U_2}{=}e\stackrel{\text{in }{U}_2}{\to }c$$give two free generators in ${\pi }_1(X,c)$. But ${\pi }_1(X,c)=\mathbb{Z}$, contradiction.

$\square$

Footnotes

1.	^ When we only consider $1$-category.