[problem:Composition Preserves Continuity | 1] Let $f: X \to Y$ and $g: Y \to Z$ be continuous maps between topological spaces. Prove that $g \circ f: X \to Z$ is continuous. [/problem]

[solution] Let $W \subseteq Z$ be open. Since $g$ is continuous, $g^{-1}(W)$ is open in $Y$. Since $f$ is continuous, $f^{-1}(g^{-1}(W))$ is open in $X$. But $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$, so $(g \circ f)^{-1}(W)$ is open. Since $W$ was arbitrary, $g \circ f$ is continuous. $\checkmark$ [/solution]

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[problem:Continuity Via The Closure Operator | 2] Prove that $f: X \to Y$ is continuous if and only if $f(\overline{A}) \subseteq \overline{f(A)}$ for every $A \subseteq X$. [/problem]

[solution] **Forward.** Assume $f$ is continuous. Let $A \subseteq X$ and $x \in \overline{A}$. We show $f(x) \in \overline{f(A)}$, i.e., every open $V$ containing $f(x)$ meets $f(A)$. Let $V \subseteq Y$ be open with $f(x) \in V$. By continuity, $U := f^{-1}(V)$ is open in $X$ with $x \in U$. Since $x \in \overline{A}$, we have $U \cap A \neq \varnothing$. Choose $z \in U \cap A$. Then $f(z) \in f(U) \cap f(A) \subseteq V \cap f(A)$. So $V \cap f(A) \neq \varnothing$, proving $f(x) \in \overline{f(A)}$. **Reverse.** Assume $f(\overline{A}) \subseteq \overline{f(A)}$ for all $A$. Let $F \subseteq Y$ be closed. Set $A = f^{-1}(F)$. Then $f(A) \subseteq F$, so $\overline{f(A)} \subseteq \overline{F} = F$ (since $F$ is closed). By hypothesis, $f(\overline{A}) \subseteq \overline{f(A)} \subseteq F$. Therefore $\overline{A} \subseteq f^{-1}(F) = A$. Since $A \subseteq \overline{A}$ always, $\overline{A} = A$, so $f^{-1}(F)$ is closed. Since $F$ was arbitrary, $f$ is continuous. $\checkmark$ [/solution]

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[problem:Product Map Continuity | 2] Let $f: X \to Y$ and $g: X \to Z$ be continuous maps. Prove that the product map $h: X \to Y \times Z$ defined by $h(x) = (f(x), g(x))$ is continuous (where $Y \times Z$ carries the product topology). [/problem]

[solution] **Step 1: Reduce to a generating set.** The product topology on $Y \times Z$ is generated by sets of the form $U \times V$ with $U$ open in $Y$ and $V$ open in $Z$. It suffices to show $h^{-1}(U \times V)$ is open in $X$ for all such sets, since a function is continuous iff preimages of a generating set are open (preimages of arbitrary unions and finite intersections are handled automatically). **Step 2: Compute the preimage.** $h^{-1}(U \times V) = \{x \in X : f(x) \in U \text{ and } g(x) \in V\} = f^{-1}(U) \cap g^{-1}(V)$. **Step 3: Apply continuity.** Since $f$ is continuous, $f^{-1}(U)$ is open in $X$. Since $g$ is continuous, $g^{-1}(V)$ is open in $X$. A finite intersection of open sets is open, so $f^{-1}(U) \cap g^{-1}(V)$ is open. $\checkmark$ This result extends to products of any family: $h: X \to \prod_\alpha Y_\alpha$ is continuous iff $\pi_\alpha \circ h$ is continuous for every $\alpha$. This is precisely the **universal property** of the product (initial) topology. [/solution]

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[problem:Continuous Image Of A Connected Space Separates No Codomain | 3] Let $X$ be a connected topological space and $f: X \to Y$ continuous. Prove that if $Y = U \cup V$ with $U, V$ open, $U \cap V = \varnothing$, and $f(X) \cap U \neq \varnothing$, then $f(X) \subseteq U$ (so $f(X) \cap V = \varnothing$). Deduce that a continuous function from a connected space to a discrete space is constant. [/problem]

[solution] **Step 1: Pull back the disconnection.** Set $A = f^{-1}(U)$ and $B = f^{-1}(V)$. Since $f$ is continuous and $U, V$ are open, both $A$ and $B$ are open in $X$. Since $Y = U \cup V$, every $x \in X$ satisfies $f(x) \in U$ or $f(x) \in V$, so $X = A \cup B$. Since $U \cap V = \varnothing$, $A \cap B = \varnothing$. **Step 2: Use connectedness.** $X = A \cup B$ with $A, B$ open and disjoint. Since $X$ is connected, one of $A, B$ must be empty. We are given $f(X) \cap U \neq \varnothing$, so $A = f^{-1}(U) \neq \varnothing$. Therefore $B = \varnothing$, meaning $f^{-1}(V) = \varnothing$, i.e., $f(X) \cap V = \varnothing$ and $f(X) \subseteq U$. **Step 3: Deduce the corollary.** If $Y$ is discrete, every singleton $\{y\}$ is open. For any $y_0 \in f(X)$, take $U = \{y_0\}$ and $V = Y \setminus \{y_0\}$. Both are open (in the discrete topology), disjoint, and $Y = U \cup V$. By Step 2, $f(X) \subseteq U = \{y_0\}$. So $f$ is constant. $\checkmark$ This gives a clean proof that connected spaces have no continuous surjection to a discrete space with two or more points, and it underlies the topological proof of the Intermediate Value Theorem: the image $f([a,b])$ is connected in $\mathbb{R}$, hence an interval. [/solution]

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[problem:The Closed Map Lemma And Quotients | 3] Let $G$ be a compact topological group acting continuously on a compact Hausdorff space $X$. Prove that the quotient map $q: X \to X/G$ (where $X/G$ carries the quotient topology) is a closed map, and that $X/G$ is Hausdorff. [/problem]

[solution] **Step 1: The quotient map is closed.** Let $C \subseteq X$ be closed. We must show $q(C)$ is closed in $X/G$, i.e., $q^{-1}(q(C))$ is closed in $X$ (by definition of the quotient topology). Now $q^{-1}(q(C)) = G \cdot C = \{g \cdot x : g \in G,\, x \in C\}$. This is the image of $G \times C$ under the action map $\mu: G \times X \to X$. Since $G$ is compact and $C$ is closed in the compact space $X$, $C$ is compact. The product $G \times C$ is compact (product of compact spaces). The action $\mu$ is continuous, so by the [Continuous Image of Compact Sets](/theorems/305), $\mu(G \times C) = G \cdot C$ is compact in $X$. Since $X$ is Hausdorff, compact subsets are closed. So $q^{-1}(q(C)) = G \cdot C$ is closed, proving $q(C)$ is closed. **Step 2: $X/G$ is Hausdorff.** Let $[x], [y] \in X/G$ with $[x] \neq [y]$, i.e., the orbits $Gx$ and $Gy$ are disjoint. Since $Gx$ is compact (image of $G$ under $g \mapsto g \cdot x$) and $Gy$ is compact, and $X$ is Hausdorff, there exist disjoint open sets $U \supseteq Gx$ and $V \supseteq Gy$. The saturations $\tilde{U} = q^{-1}(q(U))$ and $\tilde{V} = q^{-1}(q(V))$ are open in $X$ (since $q$ is an open map for group actions — the saturation of an open set under a continuous group action is open), and $q(\tilde{U})$ and $q(\tilde{V})$ separate $[x]$ and $[y]$ in $X/G$. More carefully: the complement $X \setminus U$ is closed, so $q(X \setminus U) = (X \setminus U)/{\sim}$ is closed by Step 1. Then $q(U) \supseteq X/G \setminus q(X \setminus U)$, which is open. A symmetric argument works for $V$. Since the saturations of $U$ and $V$ under the $G$-action may overlap, we refine: replace $U$ by $\bigcap_{g \in G} g \cdot U$... Actually, a cleaner approach: since $q$ is a closed surjection from a compact space to $X/G$, and closed surjections from normal spaces produce normal quotients (and compact Hausdorff spaces are normal), $X/G$ is normal, hence Hausdorff. Alternatively, use the fact that the equivalence relation $x \sim y \iff Gx = Gy$ has closed graph in $X \times X$ (the graph is $\{(x, g \cdot x) : x \in X, g \in G\} = \mu_{\text{twist}}(G \times X)$, a continuous image of a compact set, hence closed), and a quotient by an equivalence relation with closed graph in a compact Hausdorff space is Hausdorff. $\checkmark$ [/solution]