MAS61015 Algebraic Topology 4 Homeomorphism 6 Interlude on categories and functors

5. Paths

Video (Definition 5.1 to Definition 5.10)

Definition 5.1.

Let $X$ be a topological space. A path in $X$ is a continuous function $u\colon[0,1]\to X$ . If $u(0)=a$ and $u(1)=b$ then we say that $u$ is a path from $a$ to $b$ and write $u\colon a\rightsquigarrow b$ .

Example 5.2.

This picture shows a path $u$ from the point $a=(2,0)$ to the point $b=(0,-2)$ in the space $X=\{(x,y)\in{\mathbb{R}}^{2}\;|\;1\leq\|(x,y)\|\leq 3\}$ :

Explicitly, the formula is

u(t)=2(1+t-t^{2})\,\left(\cos(3\pi t/2),\;\sin(3\pi t/2)\right).

Example 5.3.

Note, however, that pictures like the one above can be a little misleading, because they only show the points traversed by the path, not the time at which those points are reached. Consider the maps $u,v\colon[0,1]\to{\mathbb{R}}^{2}$ given by $u(t)=(t,1-t)$ and $v(t)=(t^{2},1-t^{2})$ . The resulting tracks are just the same:

However, if we add markers for $t=0.1,0.2,\dotsc,0.9$ then we can see the difference:

Example 5.4.

Suppose that $X$ is a subset of ${\mathbb{R}}^{n}$ , with the subspace topology. For any points $a,b\in X$ , we can define $u\colon[0,1]\to{\mathbb{R}}^{n}$ by $u(t)=(1-t)a+tb$ . This gives a path from $a$ to $b$ in ${\mathbb{R}}^{n}$ , which we call a straight line path. However, this path might or might not lie in $X$ ; in any case where we want to use straight line paths, we need to check this. For example, if $X$ is a circle then the straight line path from $a$ to $b$ is not contained in $X$ , except in the trivial case where $a=b$ . In the space $Y$ shown on the right below, the straight line paths from $c$ to $d$ and from $d$ to $e$ are contained in $Y$ , but the straight line path from $c$ to $e$ is not contained in $Y$ .

Remark 5.5.

Later, we will want to consider paths as continuous maps $\Delta_{1}\to X$ rather than continuous maps $[0,1]\to X$ . We will always identify the point $t\in[0,1]$ with the point $(1-t,t)\in\Delta_{1}$ . This ensures that the point $0\in[0,1]$ gets identified with $e_{0}=(1,0)\in\Delta_{1}$ , and the point $1\in[0,1]$ gets identified with $e_{1}=(0,1)\in\Delta_{1}$ .

Definition 5.6.

Let $X$ be a topological space.

(a)

For any $a\in X$ , the constant path $c_{a}\colon[0,1]\to X$ is just given by $c_{a}(t)=a$ for all $t$ . This is a path $a\rightsquigarrow a$ .
(b)

Now suppose we have a path $u\colon a\rightsquigarrow b$ . We define a path $\overline{u}\colon b\rightsquigarrow a$ by $\overline{u}(t)=u(1-t)$ , and we call this the reverse of $u$ .
(c)

Now suppose we also have a path $v\colon b\rightsquigarrow c$ . We define a path $u*v\colon a\rightsquigarrow c$ by

$(u*v)(t)=\begin{cases}u(2t)&\text{ if }0\leq t\leq\tfrac{1}{2}\\ v(2t-1)&\text{ if }\tfrac{1}{2}\leq t\leq 1.\end{cases}$

(Roughly speaking, in the first half-second we go from $a$ to $b$ by following $u$ at double speed, then in the next half-second we go from $b$ to $c$ by following $v$ at double speed.)

Remark 5.7.

There are a number of things that we need to check in order to validate the above definitions, especially part (c). Firstly, $c_{a}$ is continuous by Example 3.22, and $\overline{u}$ is continuous by Proposition 3.24, because it is the composite of $u$ with the continuous function $t\mapsto 1-t$ .

Now consider $u*v$ . Firstly, if $t=\tfrac{1}{2}$ then both clauses in the definition of $(u*v)(t)$ are applicable. This would be a problem if the two clauses gave different answers for the value of $(u*v)(\tfrac{1}{2})$ . However, the first clause gives the answer $u(1)$ , and the second clause gives the answer $v(0)$ . As $u\colon a\rightsquigarrow b$ and $v\colon b\rightsquigarrow c$ we have $u(1)=b=v(0)$ so the two answers are the same.

Next, we need to show that the map $u*v\colon[0,1]\to X$ is continuous. We can write $[0,1]$ as $F_{1}\cup F_{2}$ , where $F_{1}=[0,\tfrac{1}{2}]$ and $F_{2}=[\tfrac{1}{2},0]$ . By closed patching (Proposition 3.35), it will be enough to show that $u*v$ is continuous on $F_{1}$ and also continuous on $F_{2}$ . On $F_{1}$ , we have $(u*v)(t)=u(2t)$ . This is the composite of $u$ with the map $t\mapsto 2t$ , and both of these maps are continuous, so $u*v$ is continuous by Proposition 3.24. A similar argument shows that $u*v$ is continuous on $F_{2}$ , so it is continuous on $[0,1]$ as required. We also have $(u*v)(0)=u(2\times 0)=u(0)=a$ and $(u*v)(1)=v(2\times 1-1)=v(1)=c$ , so $u*v\colon a\rightsquigarrow c$ .

Definition 5.8.

We introduce a relation on $X$ by declaring that $a\sim b$ iff there exists a path $u\colon a\rightsquigarrow b$ in $X$ .

Proposition 5.9.

This relation is an equivalence relation.

(If you need to review the basic ideas about equivalence relations, you can look forward to Definition 7.13.)

Proof.

We must show that the relation is reflexive, symmetric and transitive. In more detail, the conditions are as follows:

(a)

For all $a\in X$ , we must have $a\sim a$ .
(b)

For all $a,b\in X$ with $a\sim b$ , we must have $b\sim a$ .
(c)

For all $a,b,c\in X$ with $a\sim b$ and $b\sim a$ we must have $a\sim c$ .

For condition (a), we always have $c_{a}\colon a\rightsquigarrow a$ , and this shows that $a\sim a$ . For condition (b), suppose that $a\sim b$ . This means that there exists a path $u\colon a\rightsquigarrow b$ . It follows that the reverse path $\overline{u}$ has $\overline{u}\colon b\rightsquigarrow a$ , and thus that $b\sim a$ . Finally, suppose that $a\sim b$ and $b\sim c$ . This means that there exist paths $u\colon a\rightsquigarrow b$ and $v\colon b\rightsquigarrow c$ . The joined path $u*v$ then goes from $a$ to $c$ , proving that $a\sim c$ as required. ∎

Definition 5.10.

The equivalence classes for $\sim$ are called the path components of $X$ . We write $[a]$ for the path component containing $a$ , so that $[a]=[b]$ iff $a\sim b$ . We write $\pi_{0}(X)$ for the quotient set $X/\sim$ , or equivalently, the set of path components. We say that $X$ is path connected if it has precisely one path component. This means that $X\neq\emptyset$ , and $a\sim b$ for all $a,b\in X$ .

Video (Example 5.11 to Proposition 5.14)

Example 5.11.

Let $X$ be a subset of ${\mathbb{R}}^{n}$ . We say that $X$ is convex if for all $a,b\in X$ and $t\in[0,1]$ we have $(1-t)a+tb\in X$ . Equivalently, this means that any straight line path with endpoints in $X$ is contained wholly in $X$ , and so counts as a path in $X$ between those endpoints. It therefore follows that $X$ is path connected (provided that it is not empty).

In particular, the ball $B^{n}$ , the cube $[0,1]^{n}$ and the simplex $\Delta_{n}$ are all nonempty and convex, so they are path connected.

Example 5.12.

Consider again the following space $Y$ , which we discussed in Section 2:

This is the disjoint union of the three subsets $A$ , $B$ and $C$ . The set $A$ is the set of all points that can be connected to $a_{0}$ by a continuous path in $Y$ , or in other words $A=[a_{0}]\in\pi_{0}(Y)$ . Because $a_{0}$ , $a_{1}$ and $a_{2}$ all lie in $A$ they can all be connected to each other, which means that $a_{0}\sim a_{1}\sim a_{2}$ and $[a_{0}]=[a_{1}]=[a_{2}]=A$ . In the same way, we have $B=[b_{0}]\in\pi_{0}(Y)$ and $C=[c_{0}]=[c_{1}]\in\pi_{0}(Y)$ . From this we see that $\pi_{0}(Y)=\{A,B,C\}$ and so $|\pi_{0}(Y)|=3$ .

Remark 5.13.

You should not be confused by the fact that $A$ itself is an infinite set. The whole set $A$ taken as a single object is an element of $\pi_{0}(Y)$ , the whole set $B$ taken as a single object is another element, and the whole set $C$ is the third element.

Proposition 5.14.

For $n>0$ , the sphere $S^{n}$ is path connected.

Proof.

Suppose we have points $a,b\in S^{n}$ . Suppose for the moment that they are not opposite points, so $b\neq-a$ . Consider the linear path $u\colon[0,1]\to{\mathbb{R}}^{n+1}$ given by $u(t)=(1-t)a+tb$ . Because $a$ and $b$ are not opposite, we see that the straight line from $a$ to $b$ does not pass through the origin, so $u(t)$ is never zero. It is therefore legitimate to define $\widehat{u}(t)=u(t)/\|u(t)\|$ . This gives a continuous map $\widehat{u}\colon[0,1]\to S^{n}$ with $\widehat{u}(0)=a/\|a\|=a$ and $\widehat{u}(1)=b/\|b\|=b$ , so $a\sim b$ in $S^{n}$ . Now consider the exceptional case where $b=-a$ . If we allowed the case $n=0$ then $S^{n}$ would just consist of two points, but we have specified that $n>0$ , so $S^{n}$ is infinite, so we can choose a point $c$ that is different from $a$ and $b=-a$ . The ordinary case now tells us that $a\sim c$ and $b\sim c$ , and $\sim$ is an equivalence relation so $a\sim b$ as required.

∎

Video (Proposition 5.15 to Proposition 5.18)

Proposition 5.15.

Let $a$ and $b$ be points in a topological space $X$ . Suppose that there is a continuous function $f\colon X\to{\mathbb{R}}$ such that

(a)

$f(x)\neq 0$ for all $x\in X$ .
(b)

$f(a)<0<f(b)$ .

Then $a\not\sim b$ .

Proof.

Suppose (for a contradiction) that there is a path $u$ from $a$ to $b$ in $X$ . We then have a continuous map $g=f\circ u\colon[0,1]\to{\mathbb{R}}$ with $g(0)=f(a)<0$ and $g(1)=f(b)>0$ . By the Intermediate Value Theorem, there exists $t_{0}\in[0,1]$ with $g(t_{0})=0$ . This means that the point $x=u(t_{0})\in X$ satisfies $f(x)=0$ , contradicting assumption (a). We therefore conclude that no such path $u$ can exist, so $a\not\sim b$ . ∎

Example 5.16.

Consider the space ${\mathbb{Z}}$ . If $a,b\in{\mathbb{Z}}$ with $a<b$ , we can define $f\colon{\mathbb{Z}}\to{\mathbb{R}}$ by $f(x)=x-b+\tfrac{1}{2}$ . This is nonzero for all $x\in{\mathbb{Z}}$ , and satisfies $f(a)<0<f(b)$ , so $a\not\sim b$ . It follows that the connected components are just the singleton sets $[a]=\{a\}$ for all $a\in{\mathbb{Z}}$ , so $\pi_{0}({\mathbb{Z}})$ is essentially the same as ${\mathbb{Z}}$ .

Example 5.17.

Consider the space $GL_{2}({\mathbb{R}})$ of invertible $2\times 2$ matrices, and the elements $I,J\in GL_{2}({\mathbb{R}})$ , where

I=\left[\begin{matrix}1&0\\ 0&1\end{matrix}\right]\hskip 40.0ptJ=\left[\begin{matrix}0&1\\ 1&0\end{matrix}\right].

We can define a continuous map $f\colon GL_{2}({\mathbb{R}})\to{\mathbb{R}}$ by $f(A)=\det(A)$ . It is a standard fact of linear algebra that invertible matrices have nonzero determinant, so $f$ is nonzero everywhere on $GL_{2}({\mathbb{R}})$ . We also have $f(J)=-1$ and $f(I)=1$ . Proposition 5.15 therefore tells us that $J\not\sim I$ . More generally, we could take any $n>0$ and put

	$\displaystyle U$	$\displaystyle=\{A\in GL_{n}({\mathbb{R}})\;\|\;\det(A)>0\}$
	$\displaystyle V$	$\displaystyle=\{B\in GL_{n}({\mathbb{R}})\;\|\;\det(B)<0\}.$

The same line of argument shows that if $B\in V$ and $A\in U$ then $B\not\sim A$ . However, it can also be shown that the subsets $U$ and $V$ are both path connected (we will not give the proof here). Assuming this, we see that $U$ and $V$ are precisely the path components of $GL_{n}({\mathbb{R}})$ , so $\pi_{0}(GL_{n}({\mathbb{R}}))=\{U,V\}$ and $|\pi_{0}(GL_{n}({\mathbb{R}}))|=2$ .

Proposition 5.18.

Suppose that $X$ can be written as $X=U\cup V$ , where $U$ and $V$ are open subsets of $X$ with $U\cap V=\emptyset$ . Then for $a\in U$ and $b\in V$ we have $a\not\sim b$ .

Proof.

We define $f\colon X\to{\mathbb{R}}$ by $f(x)=-1$ for all $x\in U$ and $f(x)=1$ for all $x\in V$ . (This defines $f(x)$ for all $x$ , because $X=U\cup V$ . There is no clash between the two clauses, because $U\cap V=\emptyset$ .) We claim that $f$ is continuous. (Assuming this, the main claim follows by Proposition 5.15.) Consider an open subset $A\subseteq{\mathbb{R}}$ ; we must show that $f^{-1}(A)$ is open in $X$ . We have

f^{-1}(A)=\begin{cases}\emptyset&\text{ if }-1\not\in A\text{ and }1\not\in A% \\ U&\text{ if }-1\in A\text{ and }1\not\in A\\ V&\text{ if }-1\not\in A\text{ and }1\in A\\ X&\text{ if }-1\in A\text{ and }1\in A.\end{cases}

In all cases, we see that $f^{-1}(A)$ is open, as required. ∎

Video (Lemma 5.19 to Proposition 5.20)

Lemma 5.19.

Let $f\colon X\to Y$ be a continuous map between topological spaces. Suppose that $a,b\in X$ with $a\sim b$ , so that $[a]=[b]$ in $\pi_{0}(X)$ . Then we also have $f(a)\sim f(b)$ , and so $[f(a)]=[f(b)]$ in $\pi_{0}(Y)$ .

Proof.

We are assuming that $a\sim b$ , which means that there exists a continuous function $u\colon[0,1]\to X$ with $u(0)=a$ and $u(1)=b$ . Put $v=f\circ u$ , so $v$ is a continuous function from $[0,1]$ to $Y$ . It has $v(0)=f(u(0))=f(a)$ and $v(1)=f(u(1))=f(b)$ , so $v\colon f(a)\rightsquigarrow f(b)$ . As there exists a path in $Y$ from $f(a)$ to $f(b)$ , we have $f(a)\sim f(b)$ as claimed. ∎

Proposition 5.20.

Let $f\colon X\to Y$ be a continuous map between topological spaces. Then there is a well-defined map $f_{*}\colon\pi_{0}(X)\to\pi_{0}(Y)$ , given by $f_{*}[a]=[f(a)]$ for all $a\in X$ . Moreover:

•

For the identity map $\operatorname{id}\colon X\to X$ , we have $\operatorname{id}_{*}=\operatorname{id}\colon\pi_{0}(X)\to\pi_{0}(X)$ .
•

For any pair of continuous maps $f\colon X\to Y$ and $g\colon Y\to Z$ , we have $(g\circ f)_{*}=g_{*}\circ f_{*}\colon\pi_{0}(X)\to\pi_{0}(Z)$ .

Proof.

Let $u$ be an element of $\pi_{0}(X)$ ; we need to define $f_{*}(u)$ . We can choose a point $a\in X$ such that $u=[a]$ , and we want to define $f_{*}(u)=[f(a)]$ . The only problem with this is that it seems to depend on the choice of $a$ . If we chose a different element $b$ with $u=[b]$ , then we would also want to define $f_{*}(u)=[f(b)]$ , and that would be inconsistent if $[f(a)]$ was different from $[f(b)]$ . However, Lemma 5.19 tells us that $[f(a)]=[f(b)]$ , so this problem does not arise, and we have a well-defined function as claimed. We also have $\operatorname{id}_{*}[a]=[\operatorname{id}(a)]=[a]$ , so $\operatorname{id}_{*}$ is the identity map. Similarly, if $g\colon Y\to Z$ is another continuous map, we have

g_{*}(f_{*}([a]))=g_{*}[f(a)]=[g(f(a))]=[(g\circ f)(a)]=(g\circ f)_{*}[a],

so $g_{*}\circ f_{*}=(g\circ f)_{*}$ . ∎