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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.

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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*}

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  • $\begingroup$ @TRmate My pleasure! $\endgroup$ – triple_sec Sep 29 '15 at 16:18

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