# 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.

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