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...
please help :)
 A: 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 $\tau_p = \tau$
A: 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.
