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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?

hengxin
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1 Answers1

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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.

Najib Idrissi
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  • Could you please explain a bit more why $A \to X \times Y$ and $X \times Y \to A$ are continuous? Are you using the "Maps into Products" theorem? Thanks. – hengxin Sep 25 '14 at 12:44
  • Yes, I'm using the theorem. I'm looking at the compositions of the identity with the projections; these are continuous, therefore the map is continuous. – Najib Idrissi Sep 25 '14 at 12:46
  • I updated my answer, see if it is clearer. – Najib Idrissi Sep 25 '14 at 12:53
  • In the argument of the continuity of $A \to X \times Y$: Why is $f_1: A (= X \times Y) \to X$ (also $f_2$) continuous? I know that if $A$ is the product topology on $X \times Y$, $f_1$ is continuous. However, I think here $A = (X \times Y)$ is an any topology $\tau$ on $X \times Y$. Is any projection function $f_1: A = (X \times Y) \to X$ continuous? – hengxin Sep 25 '14 at 13:02
  • The identity $A \to A$ is continuous by definition. But a map $h: Z \to A$ is continuous if and only if $h_1$ and $h_2$ are continuous; apply this to $h = id_A : A \to A$. – Najib Idrissi Sep 25 '14 at 13:05
  • So, are you using the "Maps into Products" theorem again to prove any projection is continuous? Does this argument rely on the assumption that $A$ is associated with the product space, which is what we are trying to prove? (Still confused. Sorry for that.) – hengxin Sep 25 '14 at 13:20
  • Yes, I'm using it again, actually. I should have made that clearer. But all I'm using is that $A$ is a topological space equipped with maps $A \to X$ and $A \to Y$ that make the theorem true (ie. any map $Z \to A$ is continuous iff $Z \to A \to X$ and $Z \to A \to Y$ are continuous). – Najib Idrissi Sep 25 '14 at 13:24
  • Let us continue this discussion in chat. – hengxin Sep 25 '14 at 13:48