[proofplan] We prove the categorical epimorphism condition directly. Given continuous maps $f,g: X \to Y$ into a [Hausdorff space](/page/Hausdorff%20Space) $Y$ that agree on the dense subspace $A$, we consider the equalizer set $E = \{x \in X : f(x)=g(x)\}$. The Hausdorff property makes this equalizer closed, and density forces every closed subset of $X$ containing $A$ to be all of $X$. Therefore $E=X$, so $f=g$. [/proofplan] [step:Reduce epimorphism to equality of maps that agree on the dense subspace] Let $Y$ be a Hausdorff topological space, and let $f,g: X \to Y$ be continuous maps such that $f \circ i = g \circ i$. Since $i: A \to X$ is the inclusion map, this condition means that for every $a \in A$, \begin{align*} f(a) = g(a). \end{align*} We must prove that $f=g$ as maps from $X$ to $Y$, meaning that for every $x \in X$, \begin{align*} f(x)=g(x). \end{align*} [guided] To prove that $i$ is an epimorphism in $\mathbf{Haus}$, we use the definition of epimorphism in a category: a morphism $i: A \to X$ is an epimorphism if, for every object $Y$ and every pair of morphisms $f,g: X \to Y$, the implication \begin{align*} f \circ i = g \circ i \implies f=g \end{align*} holds. Here the objects are Hausdorff topological spaces and the morphisms are continuous maps. Thus let $Y$ be a Hausdorff topological space, and let $f,g: X \to Y$ be continuous maps satisfying $f \circ i = g \circ i$. Because $i$ is the inclusion map, $i(a)=a$ for every $a \in A$. Therefore \begin{align*} f(a) = f(i(a)) = (f \circ i)(a) = (g \circ i)(a) = g(i(a)) = g(a) \end{align*} for every $a \in A$. The task is to upgrade equality on the [dense subset](/page/Dense%20Subset) $A$ to equality on all of $X$. [/guided] [/step] [step:Show the equalizer of the two maps is closed] Define the equalizer set $E \subset X$ by \begin{align*} E := \{x \in X : f(x)=g(x)\}. \end{align*} We prove that $E$ is closed in $X$. Let $x \in X \setminus E$. Then $f(x) \neq g(x)$. Since $Y$ is Hausdorff, there exist open sets $U,V \subset Y$ such that \begin{align*} f(x) \in U, \qquad g(x) \in V, \qquad U \cap V = \varnothing. \end{align*} Because $f$ and $g$ are continuous, the sets $f^{-1}(U)$ and $g^{-1}(V)$ are open in $X$. Define the open neighbourhood $W \subset X$ of $x$ by \begin{align*} W := f^{-1}(U) \cap g^{-1}(V). \end{align*} For every $w \in W$, we have $f(w) \in U$ and $g(w) \in V$, so $f(w) \neq g(w)$ because $U \cap V=\varnothing$. Hence $W \subset X \setminus E$. Thus every point of $X \setminus E$ has an open neighbourhood contained in $X \setminus E$, so $X \setminus E$ is open. Therefore $E$ is closed in $X$. [guided] We introduce the set on which the two maps already agree: \begin{align*} E := \{x \in X : f(x)=g(x)\}. \end{align*} The goal is to prove $E=X$. The key topological input is that $E$ is closed when the target space $Y$ is Hausdorff. Let $x \in X \setminus E$. By definition of $E$, this means $f(x) \neq g(x)$ in $Y$. Since $Y$ is Hausdorff, distinct points in $Y$ can be separated by disjoint open neighbourhoods. Therefore there exist open sets $U,V \subset Y$ such that \begin{align*} f(x) \in U, \qquad g(x) \in V, \qquad U \cap V = \varnothing. \end{align*} Now we pull these separating neighbourhoods back to $X$. Since $f: X \to Y$ and $g: X \to Y$ are continuous, the preimages $f^{-1}(U)$ and $g^{-1}(V)$ are open subsets of $X$. Define \begin{align*} W := f^{-1}(U) \cap g^{-1}(V). \end{align*} Then $W$ is open in $X$ because it is the intersection of two open subsets of $X$, and $x \in W$ because $f(x)\in U$ and $g(x)\in V$. For every $w \in W$, the definitions of the two preimages give \begin{align*} f(w) \in U, \qquad g(w) \in V. \end{align*} Since $U$ and $V$ are disjoint, these two values cannot be equal. Thus $w \notin E$, and therefore $W \subset X \setminus E$. We have shown that each point of $X \setminus E$ has an open neighbourhood contained in $X \setminus E$, so $X \setminus E$ is open. Hence $E$ is closed in $X$. [/guided] [/step] [step:Use density to force the closed equalizer to be all of $X$] From $f \circ i = g \circ i$, we have $f(a)=g(a)$ for every $a \in A$, so $A \subset E$. Since $A$ is dense in $X$, its closure in $X$ satisfies \begin{align*} \overline{A}=X. \end{align*} Because $E$ is closed in $X$ and contains $A$, it contains $\overline{A}$. Therefore \begin{align*} X=\overline{A} \subset E \subset X. \end{align*} Hence $E=X$. [guided] We now combine the two ingredients: agreement on $A$ and closedness of the equalizer. From the assumption $f \circ i=g \circ i$, and because $i$ is the inclusion map, we proved that \begin{align*} f(a)=g(a) \end{align*} for every $a \in A$. By the definition of $E$, this says exactly that $A \subset E$. The subset $A$ is dense in $X$, so its closure in $X$ is all of $X$: \begin{align*} \overline{A}=X. \end{align*} The set $E$ is closed in $X$ by the previous step. Any closed subset of $X$ that contains $A$ must contain the closure of $A$, because $\overline{A}$ is the smallest closed subset of $X$ containing $A$. Therefore \begin{align*} X=\overline{A} \subset E. \end{align*} Since $E$ was defined as a subset of $X$, we also have $E \subset X$. Thus \begin{align*} X=\overline{A} \subset E \subset X, \end{align*} and consequently $E=X$. [/guided] [/step] [step:Conclude that the inclusion is an epimorphism in $\mathbf{Haus}$] Since $E=X$, for every $x \in X$ we have $x \in E$, and hence \begin{align*} f(x)=g(x). \end{align*} Thus $f=g$ as maps $X \to Y$. Since this holds for every Hausdorff space $Y$ and every pair of continuous maps $f,g: X \to Y$ satisfying $f \circ i=g \circ i$, the inclusion map $i: A \to X$ is an epimorphism in $\mathbf{Haus}$. [/step]