The inclusion function $Haus\hookrightarrow Top$ has a left adjoint.

Question

Can someone help me with my proof that the inclusion functor $Haus\hookrightarrow Top$ has a left adjoint? The inclusion functor is continuous. However, I am failing to prove that the inclusion functor satisfies the solution set condition.

Answer 1

Define the relation $R$ on every topological $X$ such that $xRy$ iff every open subset which contains $x$ contains $y$ and $S$ the equivalence relation generated by $R$. We denote by $X_S$ the set of equivalence classes with the quotient topology. The map $X\rightarrow X_S$ is the right adjoint.

Answer 2

Let $X$ be any topological space. Since $\textbf{Haus}$ is a full subcategory of $\textbf{Top}$, it is enough to provide a set $\{X_i\}_{i \in I}$ of Hausdorff spaces such that every continuous map $f: X \to Y$, where $Y$ is Hausdorff, factors through some $X_i$. We can take $\{X_i\}_{i \in I}$ to be the set of Hausdorff quotients of $X$.

Let $f: X \to Y$ be continuous, with $Y$ Hausdorff. Define an equivalence relation $R$ on $X$ by $x R x'$ iff $f(x) = f(x')$, and let $\pi: X \to X / R$ be the quotient map. Then $f$ factors as $\bar{f} \pi$, for some continuous $\bar{f}: X/R \to Y$. So we are left to show that $X / R$ is Hausdorff. Let $\pi(x), \pi(x') \in X/R$ be distinct, then by definition of $R$ we have $$ \bar{f}(\pi(x)) = f(x) \neq f(x') = \bar{f}(\pi(x')). $$ So by Hausdorffness of $Y$, we find opens $\bar{f}(\pi(x)) \in V$ and $\bar{f}(\pi(x')) \in V'$, such that $V \cap V' = \emptyset$. Then $\bar{f}^{-1}(V)$ and $\bar{f}^{-1}(V')$ are opens in $X / R$, containing $\pi(x)$ and $\pi(x')$ respectively, while having empty intersection. So indeed, $X / R$ is Hausdorff.

Edit: as mentioned in the comments by Jeremy, a more elegant solution would be to take $\{X_i\}_{i \in I}$ to be the set of all Hausdorff spaces with a surjective continuous map from $X$. Since every continuous map factors as a surjective map (to a subspace) followed by an injective map (the inclusion of that subspace), this would work as well.