TITLE: What is the "Tychonoffication"?
QUESTION [6 upvotes]: In this link:
https://mathoverflow.net/questions/23940/why-free-topological-groups-on-tychonoff-spaces
I read the following:
Let $X$ be a topological space. The Tychonoffication $Y$ of $X$ is the quotient of $X$ by the relation $x\sim y$ iff $f(x)=f(y)$ for all continuous $f:X\to\mathbb{R}$, and we give $Y$ the weak topology induced by all these real-vauled maps.
This makes $Y$ a Tychonoff  space that satisfies the universal property: any continuous map from $X$ to a Tychonoff space factors uniquely through $Y$.
I don't understand two things:
1) The weak topology of a family of functions $f_i:X\to X_i$ is the topology over $X$ which has the subbase $\{f^{-1}_i(U):...\}$. How can the real-valued functions $f:X\to\mathbb{R}$ give a topology over $Y$?
2) What does it mean "factors through $Y$"?
Thanks.

REPLY [2 votes]: 1) I think it means the following: For any continuous function $f:X\to\mathbb R$, there corresponds a function $F_f:Y\to\mathbb R$, defined as $F_f([x])\equiv f(x)$, where $x\in X$ is an arbitrary member of the equivalence class $[x]\in Y$. The quantity $F_f([x])$ is well-defined and does not depend on the particular element $x$ of the equivalence class, because all elements of the equivalence class give the same value for $f$ by the definition of the equivalence relation $\sim$. Now, given this family of real-valued functions $(F_f)_{f\in\mathbb R^X\text{ continuous}}$, one can define a “weak topology” on $Y$ as the topology generated by the following subbase: $$\{F_f^{-1}(U)\,|\,f\in\mathbb R^X\text{ is continuous and }U\subseteq\mathbb R\text{ is open}\}.$$

2) Let $\pi:X\to Y$ be the quotient map: if $x\in X$, then $\pi(x)\equiv[x]$ is defined to be the equivalence class $x$ is in. Moreover, let $Z$ be any Tychonoff space and let $h:X\to Z$ be a continuous function. Then, that “$h$ uniquely factors through $Y$” means that there exists a unique function $g:Y\to Z$ such that

$g$ is continuous (with respect to the weak topology on $Y$ constructed above); and
$h=g\circ \pi$;

that is, the following diagram commutes:
\begin{align*}
\begin{array}{ccl}
X&\overset{\pi}{\longrightarrow}&Y&\\
&\underset{h}{{\searrow}}&\big{\downarrow}\scriptsize{g}\\
&&Z&
\end{array}
\end{align*}