Does box topology satisfy $f:Y \to \prod X_i$ is continuous iff the composition $p_if$ is continuous

Question

In my text, it is stated that the following definition below is another way of defining the product topology:

i) for each factor, $p_i$ is continuous, ii) A function $f:Y \to \prod X_i$ is continuous iff the composition $p_if$ is continuous for all $i$

They two conditions don't look that restrictive so I was wondering why this is not true for the box topology. Certainly, i) is satisfied so the problem must be with ii). The forward direction is certainly satisfied and there is no difference with the product topology. So I guess the problem must be with the backward direction right?

Answer 1

The box topology does not satisfy that criterion.

Let $X= \Bbb R^{\Bbb N}$ in the box topology (so countably many copies of $\Bbb R$ in the standard topology).

Then $f: \Bbb R \to X$ defined by $f(x)=(x,x,x,\ldots)$ so that $p_n \circ f = \textrm{id}$ fulfills the right hand side of ii, but $f$ is not continuous as

$f^{-1}[\prod_n (-\frac1n, \frac1n)] = \{0\}$ is not open in $\Bbb R$.

The product topology has the handy property that you can conclude the continuity of $f$ going into a product, just from its compositions with all projections, and it's the unique topology on the product with that property. This is a common thing in so-called initial topologies.