4.6 Continuous Maps
Definition 4.6.1 (Continuity). Let $X$ and $Y$ be topological spaces, $f: X \to Y$ be a function, and $x \in X$, then the following are equivalent:
For each $V \in \cn(f(x))$, $f^{-1}(V) \in \cn(x)$.
For each filter base $\fB \subset 2^{X}$ converging to $x$, $f(\fB)$ converges to $f(x)$.
If the above holds, then $f$ is continuous at $x \in X$.
The following are also equivalent:
For each $U \subset Y$ open, $f^{-1}(U)$ is open in $X$.
$f$ is continuous at every $x \in X$.
For each convergent filter base $\fB \subset 2^{X}$, $f(\fB)$ is convergent.
If the above holds, then $f$ is continuous.
The collection $C(X; Y)$ is the space of all continuous functions from $X$ to $Y$.
Proof. Local continuity, $(1) \Rightarrow (2)$: Let $V \in \cn(f(x))$, then $f^{-1}(V) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset f^{-1}(V)$. In which case, $f(B) \subset V$ and $V \in \fF(\fB)$ by (F1).
Local continuity, $(2) \Rightarrow (1)$: Since $f(\cn(x))$ converges to $f(x)$, for each $U \in \cn(f(x))$, there exists $V \in \cn(x)$ such that $f(V) \subset V$. Thus $V \subset f^{-1}(U) \in \cn(x)$ by (F1).
Global continuity, $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $U \in \cn(f(x))$ for all $x \in f^{-1}(U)$. Hence $f^{-1}(U) \in \cn(x)$ for all $x \in f^{-1}(U)$. Thus $U$ is open by Lemma 4.4.3.
Global continuity, $(2) \Rightarrow (1)$: Let $x \in X$ and $V \in \cn(f(x))$, then there exists $U \in \cn(f(x))$ open, so $f^{-1}(V) \supset f^{-1}(U)$ contains an open set that contains $x$. Therefore $f^{-1}(V) \in \cn(x)$.
Global continuity, $(2) \Leftrightarrow (3)$: Follows from local continuity.$\square$
Lemma 4.6.2 (Gluing Lemma for Continuous Functions). Let $X, Y$ be topological spaces, $\seq{U_i}\subset 2^{X}$ be open sets, and $\seqi{f}$ with $f_{i} \in C(U; Y)$ for all $i \in I$. If:
$\bigcup_{i \in I}U_{i} = X$.
For each $i, j \in I$, either $U_{i} \cap U_{j} = \emptyset$, or $f_{i}|_{U_i \cap U_j}= f_{j}|_{U_i \cap U_j}$.
then there exists a unique $f \in C(X; Y)$ such that $f|_{U_i}= f_{i}$ for all $i \in I$.
Proof. By Lemma 2.0.11, there exists a unique $f: X \to Y$ with $f|_{U_i}= f_{i}$ for all $i \in I$.
Let $x \in X$. By assumption (a), there exists $i \in I$ such that $x \in U_{i}$. Since $f_{i} \in C(U_{i}; Y)$, for any $V \in \cn_{Y}(f(x))$, there exists $U \in \cn_{U_i}(x)$ such that $f_{i}(U) \subset V$. As $U_{i}$ is open in $X$, $U \in \cn_{X}(x)$. Thus $f(U) = f_{i}(U) \subset V$, and $f$ is continuous at $x$.$\square$