Continuous Maps on Topological Spaces Review
Continuous Maps on Topological Spaces Review
We will now review some of the recent material regarding continuous maps on topological spaces.
Let $X$ and $Y$ be topological spaces and let $f : X \to Y$.
- Recall from the Continuous Maps on Topological Spaces page that $f$ is said to be Continuous at a Point $a \in X$ if there exists local bases $\mathcal B_a$ of $a$ and $\mathcal B_{f(a)}$ of $f(a)$ such that for all $B \in \mathcal B_{f(a)}$ there exists a $B' \in \mathcal B_a$ such that:
\begin{align} \quad f(B') \subseteq B \end{align}
- Furthermore, $f$ is said to be Continuous or more specifically, Continuous on All of $X$ if $f$ is continuous at every $a \in X$.
- On The Open Neighbourhood Definition of Continuous Maps on Topological Spaces page we looked at an equivalent definition for a map $f$ to be continuous. We said that $f$ is continuous at a point $a \in X$ if and only if for every open neighbourhood $V$ of $f(a)$ there exists an open neighbourhood $U$ of $a$ such that:
\begin{align} \quad f(U) \subseteq V \end{align}
- On the Equivalent Statements Regarding Continuous Maps on Topological Spaces page we saw that $f$ is continuous on all of $X$ implies that for all open sets $V$ in $Y$ we have that the inverse image $f^{-1}(V)$ is open in $X$.
- We then showed that if for all open sets $V$ in $Y$ we have that the inverse image $f^{-1}(V)$ is open in $X$ then this implies that for every basis $\mathcal B_Y$ of $Y$ and for every $B \in \mathcal B_Y$ we have that $f^{-1}(B)$ is open in $X$ (since every element in the basis of the topology on $Y$ is open by definition).
- Furthermore, we showed that if for every basis $\mathcal B_Y$ of $Y$ we have that for every $B \in \mathcal B_Y$ we have that $f^{-1}(B)$ is open in $X$ then this implies that $f$ is continuous on all of $X$ which completes our cycle of equivalent statements.
- On the The Closed Set Definition of Continuous Maps on Topological Spaces we went further to show that a function $f$ is continuous on all of $X$ if and only if for every closed set $V$ in $Y$ we have that $f^{-1}(V)$ is closed in $X$.
- All of these equivalent definitions of continuity of $f$ were then summarized on the Summary of Equivalent Statements Regarding Continuous Maps on Topological Spaces page.
- On the Continuity of the Composition of Continuous Maps on Topological Spaces page we saw that if $X$, $Y$, and $Z$ were all topological spaces and $f : X \to Y$ and $g : Y \to Z$ are continuous functions then the composite function $g \circ f : X \to Z$ defined for all $x \in X$ by $(g \circ f)(x) = g(f(x))$ is also inherently continuous on all of $X$.