Math4201 Topology I (Lecture 11)

Note

Q: Let $f:X\to Y$ be a continuous bijection. Is it true that $f^{-1}$ is continuous?

A: No. Consider $X=[0,2\pi)$ and $Y=\mathbb{S}^1$ with standard topology in $\mathbb{R}^2$ .

Let $f\coloneqq \theta\in [0,2\pi)\to (\cos \theta, \sin \theta)\in \mathbb{S}^1$ is a continuous bijection. ( $\forall f^{-1}(V)$ is open in $X$ )

But $f^{-1}$ is not continuous, consider the open set in $X, U=[0,\pi)$ . Then $f^{-1}(U)=[0,\pi)$ is not open in $Y$ .

Continuous functions

Constructing continuous functions

Theorem composition of continuous functions is continuous

Let $X,Y,Z$ be topological spaces, $f:X\to Y$ is continuous, and $g:Y\to Z$ is continuous. Then $f\circ g:X\to Z$ is continuous.

Proof

Let $U\subseteq Z$ be open. Then $g^{-1}(U)$ is open in $Y$ . Since $f$ is continuous, $f^{-1}(g^{-1}(U))$ is open in $X$ .

Pasting lemma

Let $X$ be a topological space and $X=Z_1\cup Z_2$ with $Z_1,Z_2$ closed in $X$ equipped with the subspace topology. (may be not disjoint)

Let $g_1:Z_1\to Y$ and $g_2:Z_2\to Y$ be two continuous maps and $\forall x\in Z_1\cap Z_2$ , $g_1(x)=g_2(x)$ .

Define $f:X\to Y$ by $f(x)\begin{cases}g_1(x), & x\in Z_1 \\ g_2(x), & x\in Z_2\end{cases}$ is continuous.

Proof

Let $U\subseteq Y$ be open. Then $f^{-1}(U)=g_1^{-1}(U)\cup g_2^{-1}(U)$ .

$g_1^{-1}(U)$ and $g_2^{-1}(U)$ are open in $Z_1$ and $Z_2$ respectively.

It’s a bit annoying to show that $g_1^{-1}(U)$ and $g_2^{-1}(U)$ are open in $X$ .

Different way. Consider the definition of continuous functions using closed sets.

If $W\subseteq X$ is closed, then $W=Z_1\cap Z_2$ is closed in $X$ .

So $f^{-1}(W)=g_1^{-1}(W)\cup g_2^{-1}(W)$ is closed in $Z_1$ and $Z_2$ respectively.

Note that $Z_1$ and $Z_2$ are closed in $X$ , so $g_1^{-1}(W)$ and $g_2^{-1}(W)$ are closed in $X$ . closed in closed subspace lemma

So $f^{-1}(W)$ is closed in $X$ .

Let $X$ be a topological space and $X=U_1\cup U_2$ with $U_1,U_2$ open in $X$ equipped with the subspace topology.

With $g_1:U_1\to Y$ and $g_2:U_2\to Y$ be two continuous maps and $\forall x\in U_1\cap U_2$ , $g_1(x)=g_2(x)$ .

Then $f:X\to Y$ by $f(x)\begin{cases}g_1(x), & x\in U_1 \\ g_2(x), & x\in U_2\end{cases}$ is continuous.

Proof

Let $U\subseteq Y$ be open. Then $f^{-1}(U)=g_1^{-1}(U)\cup g_2^{-1}(U)$ .

$g_1^{-1}(U)$ and $g_2^{-1}(U)$ are open in $U_1$ and $U_2$ respectively.

Apply the open in open subspace lemma

So $f^{-1}(U)$ is open in $X$ .

The open set version holds more generally.

Let $X$ be a topological space and $X=\bigcup_{\alpha\in I} U_\alpha$ with $U_\alpha$ open in $X$ equipped with the subspace topology.

Let $g_\alpha:U_\alpha\to Y$ be two continuous maps and $\forall x\in U_\alpha\cap U_\beta$ , $g_\alpha(x)=g_\beta(x)$ .

Then $f:X\to Y$ by $f(x)=g_\alpha(x), \text{if } x\in U_\alpha$ is continuous.

Continuous functions on different codomains

Let $f:X\to Y$ and $g:X\to Z$ be two continuous maps of topological spaces.

Let $H:X\to Y\times Z$ , where $Y\times Z$ is equipped with the product topology, be defined by $H(x)=(f(x),g(x))$ . Then $H$ is continuous.

A stronger version of this theorem is that $f:X\to Y$ and $g:X\to Z$ are continuous maps of topological spaces if and only if $H:X\to Y\times Z$ is continuous.

Proof

It is sufficient to check the basis elements of the topology on $Y\times Z$ .

The basis for the topology on $Y\times Z$ is $U\times V\subseteq Y\times Z$ , where $U\subseteq Y$ and $V\subseteq Z$ are open. This form a basis for the topology on $Y\times Z$ .

We only need to show that $H^{-1}(U\times V)$ is open in $X$ .

Let $H^{-1}(U\times V)=\{x\in X | (f(x),g(x))\in U\times V\}$ .

So $H^{-1}(U\times V)=f^{-1}(U)\cap g^{-1}(V)$ .

Since $f$ and $g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open in $X$ .

So $H^{-1}(U\times V)$ is open in $X$ .

Exercise: Prove the stronger version of the theorem,

If $H:X\to Y\times Z$ is continuous, then $f:X\to Y$ and $g:X\to Z$ are continuous.