Subcategories of T1, Hausdorff and TD spaces are reflective subcategories of Category *Top*.

Question

How do I construct a T1, Hausdorff and TD reflections to show that subcategories of T1, Hausdorff and TD spaces are reflective subcategories of Category Top?

The T0 reflection is just the T0 quotient of a space X. What about the others?

Answer 1

This question and its answers talk in detail about the Hausdorff reflection.

The $T_1$ reflection of $X$ is the quotient under the intersection of all equivalence relations on $X$ that have closed equivalence classes, see the post here with an attempted proof (the previous one has the statement).

My details: Let $X$ be any space and define $$\mathcal{C} = \{S \subseteq X \times X: S \text{ is an equivalence relation and } \forall x \in X: [x]_S \text{ closed in } X\}$$ where $[x]_S$ is the class of $x \in X$ under $S$. This is a non-empty set of equivalence relations as the trivial relation $R= X \times X$ is in $\mathcal{C}$. As the intersection of any family of equivalence relations is again an equivalence relation $R= \bigcap\{S \in \mathcal{C}\}$ is well- defined. As for any $x$ we have $$[x]_R = \bigcap \{[x]_S: S \in \mathcal{C}\}$$ and intersections of closed sets are closed, we have that $R$ also has the property that all classes are closed and so $X/R$ is a well-defined $T_1$ space in the quotient topology (with quotient map $q: X \to X/R$ defined as usual by $q(x)= [x]_R$). This $(X/R,q)$ is the $T_1$-reflection of $X$: let $Y$ be any $T_1$ space and $f: X \to Y$ continuous. Then $R_f = \{(x,x') \in X \times X: f(x)=f(x')\}$ is an equivalence relation and as $[x]_{R_f} = f^{-1}[\{f(x)\}]$, $R_f \in \mathcal{C}$, and so $R \subseteq R_f$, which implies that if $[x]_R =[x']_R$ we have that $f(x) = f(x')$ too, so that $\tilde{f}([x]_R) = f(x)$ is well-defined and continuous as $q^{-1}[\tilde{f}^{-1}[O]] = f^{-1}[O]$ for all $O$ and $X/R$ has the quotient topology w.r.t. $q$. It's clear that $\tilde{f}$ is unique.

That same linked thread claims that $T_D$ is not reflective because $S_2^{\aleph_0}$, where $S_2$ is the Sierpiński two-space, is not $T_D$. More is explained in the introduction of this paper.

Answer 2

@Henno . This is what I tried for the T1 Reflection. But I hardly used the fact that singleton sets of a T1 space are closed. Any suggestion?

Let $R$ be the smallest equivalence relation $S$ on $X$ such that $X/S$ is closed. Denote by $X/ R$ the quotient space and $p$ the quotient map from $X$ onto $X/ R$. We show that $X /R$ is the $T_{1}$ reflection of $X$. That $X/R$ is $T_{1}$ follows from the fact that a quotient space is $T_{1}$ if and only if each equivalence class of $X$ is closed, [41]. Let $f:X\longrightarrow Y$ be a continuous map from $X$ to a $T_{1}$ space $Y$. Define a function $g:X/R\longrightarrow Y$ by $[x]\mapsto f(x)$. Observe that $g$ is defined since for any $[x], [y]\in X/R$, if $g([x])\neq g([y])$, then $f(x)\neq f(y)$. Now, $\{f(x)\}$ and $\{f(y)\}$ are disjoint closed subsets of $Y$. Therefore, $f^{-1}(\{x\})$ and $f^{-1}(\{y\})$ are disjoint closed sets of $X$. Because $[x]\subseteq f^{-1}(\{f(x)\})$ and $[y]\subseteq f^{-1}(\{f(y)\})$, it follows that $[x]\neq [y]$.\ For continuity of $g$, let $U$ be a closed subset of $Y$. Then $f^{-1}(U)$ is closed in $X$. Observe that \begin{align*} g^{-1}(U)&=\{[x]:f(x)\in U\}\\ &=\{[x]:x\in f^{-1}(U)\}\\ &=p(f^{-1}(U)) \end{align*} Now, \begin{align*} p^{-1}(p(f^{-1}(U)))&=\{x: f(x)\in p(f^{-1}(U))\}\\ &=\{x:x\in f^{-1}(U)\}\\ &=f^{-1}(U)\subseteq f^{-1}(U)\subseteq p^{-1}(p(f^{-1}(U))) \end{align*} Therefore, $p^{-1}(p(f^{-1}(U)))=f^{-1}(U)$. By definition of a quotient map $p$, $g^{-1}(U)$ is closed in $X/R$. Thus $g$ is continuous.\ Clearly $g$ is unique.\ So, the $T_{1}$ quotient space $X/R$ is the $T_{1}$ reflection of $X$.