Topological Continuity

Throughout this writeup, let $K, L$ be sets and $f : K \to L$ be a function. Refer to a previous post for prior topological notions that will be used in this set of exercises.

Problem 1. For any topology $\mathcal T$ on $L$ , define $\mathcal T_{f^{-1}}$ by

$\mathcal T_{f^{-1}} := \{f^{-1}(V) : V \in \mathcal T\}.$

Prove that $\mathcal T_{f^{-1}}$ is a topology on $K$ , known as the pull-back topology of $f$ .

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Solution. We have $\emptyset = f^{-1}(\emptyset) \in \mathcal T_{f^{-1}}$ and $K = f^{-1}(L) \in \mathcal T_{f^{-1}}$ .

The union and intersection properties follows from set algebra via

$\displaystyle \begin{aligned} \bigcup_\alpha f^{-1}(V_\alpha) &= f^{-1} \left( \bigcup_\alpha V_\alpha \right) \in T_{f^{-1}}, \\ \bigcap_{i=1}^n f^{-1}(V_i) &= f^{-1} \left( \bigcap_{i=1}^n V_i \right) \in T_{f^{-1}}. \end{aligned}$

Problem 2. For any topology $\mathcal S$ on $X$ , define $\mathcal S_{f}$ by

$\mathcal S_{f} := \{V \in \mathcal P(Y) : f^{-1}(V) \in \mathcal S\}.$

Prove that $\mathcal S_{f}$ is a topology on $Y$ , known as the push-forward topology of $f$ .

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Solution. Follow the proof in Problem 1.

Let $\mathcal S$ be a topology on $K$ and $\mathcal T$ be a topology on $L$ .

Definition 1. A function $\phi : K \to L$ is said to be continuous if for any $V \in \mathcal T$ , $\phi^{-1}(V) \in \mathcal S$ .

Problem 3. Prove that $f$ is continuous if and only if $\mathcal T_{f^{-1}} \subseteq \mathcal S$ . Since this holds for any topology $\mathcal S$ , we say that $\mathcal T_{f^{-1}}$ is the coarsest topology such that $f$ is continuous.

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Solution. We first prove $(\Rightarrow)$ . Suppose $f$ is continuous. Fix $U \in \mathcal T_{f^{-1}}$ . Then $U = f^{-1}(V)$ for some $V \in \mathcal T$ , which implies $U = f^{-1}(V) \in \mathcal S$ , as required.

Next, we prove $(\Leftarrow)$ . Suppose $\mathcal T_{f^{-1}} \subseteq \mathcal S$ . Fix $V \in \mathcal T$ . By definition, $f^{-1}(V) \in \mathcal T_{f^{-1}} \subseteq \mathcal S$ . Therefore, $f$ is continuous.

Problem 4. Prove that $f$ is continuous if and only if $\mathcal T \subseteq \mathcal S_f$ . Since this holds for any topology $\mathcal T$ , we say that $\mathcal S_f$ is the finest topology such that $f$ is continuous.

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Solution. We first prove $(\Rightarrow)$ . Suppose $f$ is continuous. Fix $V \in \mathcal T$ . By continuity, $f^{-1}(V) \in \mathcal S$ , so that $V \in \mathcal S_f$ .

Next, we prove $(\Leftarrow)$ . Suppose $\mathcal T \subseteq \mathcal S_f$ . Fix $V \in \mathcal T$ . By assumption, $V \in \mathcal T \subseteq \mathcal S_f$ , so that $f^{-1}(V) \in \mathcal S$ . Therefore, $f$ is continuous.

Problem 5. For any $J \subseteq K$ , define the inclusion map by $\iota : J \to K$ , $\iota(x) = x$ . Prove that $\mathcal S_{\iota^{-1}} = \{J \cap V : V \in \mathcal S\}$ . That is, the subspace topology is the smallest topology such that the inclusion map $\iota$ is continuous.

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Solution. Each element in $\mathcal S_{\iota^{-1}}$ is of the form $\iota^{-1}(V)$ for some $V \in \mathcal S$ . Since $\iota^{-1}(V) \subseteq J$ and $\iota^{-1}(V) \subseteq V$ , we have $\iota^{-1}(V) \subseteq J \cap V$ . The result follows by noticing $\iota(J \cap V) \subseteq V \Rightarrow J \cap V \subseteq \iota^{-1}(V)$ .

Problem 6. Let $(M, \mathcal U)$ be a topological space. Suppose $g : L \to M$ is continuous. Prove that $g\circ f : K \to M$ is continuous.

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Solution. Fix $W \in \mathcal U$ . Since $g$ is continuous, $g^{-1}(W) \in \mathcal T$ . Since $f$ is continuous, $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W)) \in \mathcal S$ . Therefore, $g \circ f$ îs continuous.