# Prove $\tau$ is the product topology (Or: inverse of Id map is continuous)

Let $(X_\alpha)_{\alpha\in I}$ a collection of non-empty topological spaces.

Denote $X:=\prod_{\alpha\in I}X_\alpha$, with $\tau$ - its topology.

Given that: for every topological space $Y$ and $f:Y\to X$, ($f$ is continuous) iff ($\forall \alpha\in I: \pi_\alpha\circ f$ is continuous ).

Where $\pi_\alpha$ is the projection from X to the $\alpha$ coordinate.

Prove $\tau$ is the product topology.

## My thoughts:

Denote the product topology $\tau_p$

Choose $f=Id:(X,\tau_p)\to (X,\tau)$

Let $V_\alpha\subseteq X_\alpha$ be an open set, hence $(\pi_\alpha\circ f)^{-1}(V_\alpha)=\pi_\alpha^{-1}(V_\alpha)=V_\alpha\times\prod_{\beta\neq \alpha} X_\beta\in\tau_p$

so $Id$ is continuous and $\tau\subseteq\tau_p$.

I cannot figure the other direction, the given argument only discusses a function from Y to X, and not the other way around, so I cannot somehow conclude (can I?) that $Id:(X,\tau)\to(X,\tau_p)$ is continuous...

You have proven that $Id:(X,\tau_p)\rightarrow (X,\tau)$ is continuous, so you have proven that $\tau$ is coarser (has less opens) than $\tau_p$. (as you said, $\tau \subseteq \tau_p$)

From the given you also have that all of the projections are continuous, by just composing with $f = Id:(X,\tau)\rightarrow (X,\tau)$, because then $f$ is continuous iff $\pi_a \circ f$ is. The first one is obviously continuous, so the latter one is continuous as well and is obviously equal to $\pi_a$

Now, by using the fact (or definition) that $\tau_p$ is the coarsest topology that makes all of the projections continous, we must have $\tau_p \subseteq \tau$ and then ofcourse $\tau_p = \tau$

First note that all $\pi_\alpha$ are ($\tau$-$\tau_\alpha$)-continuous. This follows from the left to right direction of the continuity criterion applied to the identy of $(\prod_\alpha X_\alpha, \tau)$ to itself which we always know to be continuous!

Indeed consider the identity $1: (\prod_\alpha X_\alpha, \tau) \to (\prod_\alpha X_\alpha, \tau_p)$.

Then $\pi_\alpha \circ 1 = \pi_\alpha: (\prod_\alpha X_\alpha, \tau) \to X_\alpha$ is continuous for all $\alpha$ and so by the right to left direction of the continuity criterion, we know $1$ is continuous, or $\tau_p \subseteq \tau$.

But likewise the other identity $1': \prod_\alpha X_\alpha, \tau_p) \to (\prod_\alpha X_\alpha, \tau)$ is continuous as all projections are $(\tau_p, \tau_\alpha)$-continuous by the definition of the product topology. Hence $\tau \subseteq \tau_p$ from this continuity and we have equality of topologies.

• Thanks, how did you conclude the continuity of $\pi_\alpha\circ 1$ (your 4th line)? to prove this is continuous, you should show that every open set of $X\_\alpha$ has an open source, but at this point - we know nothing about the sets contained in $\tau$... what am I missing? Commented Jul 12, 2017 at 8:17
• @Daniel I already deduced in the beginning that all projections are continuous in $\tau$ and the compositions are exactly those projections. No need to know about what $\tau$ looks like; we only use the continuity criterion. Commented Jul 12, 2017 at 8:21
• you are right :) Many thanks Commented Jul 12, 2017 at 8:40