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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\coloneq \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.