The product topology is the only one on $X \times Y$ that makes the "Maps into Products" theorem valid

Question

The "Maps into Products" theorem says that,

(Maps into Products) Let $f: A \to X \times Y$ be given by the equation $$f(a) = (f_1(a), f_2(a)).$$ Then $f$ is continuous iff the functions $$f_1: A \to X \textrm{ and } f_2: A \to Y$$ are continuous.

The book from which I am learning "general topology" comments that,

Comment: Product topology is the only topology on $X \times Y$ which makes the "Maps into Products" theorem valid.

So, my question is:

Question: How to prove the comment?

My attempt: I notice that the proof of the "Maps into Products" theorem makes use of the continuity of both $\pi_1: X \times Y \to X$ and $\pi_2: X \times Y \to Y$, where $\pi_i, \pi_2$ are projection functions.
And I know that the product topology on $X \times Y$ is the minimal one which makes both $\pi_1$ and $\pi_2$ continuous.
Then how to proceed?

Answer 1

Here's a quick proof.

Suppose that there is another topology $\tau$ on $X \times Y$ that makes the theorem true, and let $A$ be the set $X \times Y$ endowed with the topology $\tau$. You also know that the theorem true for the usual product topology.

Let $f : A \to X \times Y$ be the identity map. Then the maps $f_1 : A \to X$ and $f_2 : A \to Y$ are the projections, and they are therefore continuous; it follows then that $f$ is continuous. The inverse of $f$ (still the identity) $g = f^{-1} : X \times Y \to A$ is also continuous, for the same reason ($g_1$ and $g_2$ are the projections. Hence the identity $A \to X \times Y$ is a homeomorphism, and the topology $\tau$ is the same as the product topology.