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Given $X$ and $Y$ topological spaces, I need to

prove that the product topology on $X \times Y$ is the finest one such that for any topological space $Z$, and continuous functions $f: Z \to X$ and $g: Z \to Y$, the function $(f\times g): Z\to X\times Y$ is continuous.

What I did was to write $f=\pi_1\circ(f\times g)$ and $g=\pi_2\circ(f\times g)$.

Then take an open set $U$ in the topology of X. Since $f$ is continuous,

\begin{equation} f^{-1}(U)=(f\times g)^{-1}( \pi_1^{-1}(U))=(f\times g)^{-1}(U\times Y) \end{equation}

is an open set in $Z$. Similarly, we get that $(f\times g)^{-1}(X\times V)$ is also an open set in Z, where $V$ is open in $Y$.

This means that we need the preimage of all sets of these forms to be open in $Z$, but $\{X\times V\} \cup \{U \times Y\}$ are a subbase for the product topology.

Is my reasoning correct? How do I prove there is no finest topology that works?

mattosmat
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  • @JeanMarie I'm trying to go in the opposite direction. Note that I go from $Z$ to $X\times Y$. The idea is to prove this and also that the product topology is the coarsest one that makes projections continuous and to conclude that those two properties give a unique characterization of the product topology. – mattosmat Aug 31 '16 at 16:14
  • All right, my bad... I had read too rapidly, and thought it was projections... I erase my first remark in order not to mislead further readers... – Jean Marie Aug 31 '16 at 16:21

1 Answers1

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Your reasoning is correct. One suggestion for the notation, though: You should write $$ (f\times g)^{-1}(\pi_1^{-1}(U)) $$ instead of $$ \color{red}{(f\times g)^{-1}\circ \pi_1^{-1}(U)} $$ since you are not composing any inverse functions here. Your notation would only work if $(f \times g)$ and $\pi_1$ were bijective function and thus had inverse functions which you could compose.

Here is an approach for the other part of the problem: Let $(X \times Y, \tau_p)$ be the space $X \times Y$ with the product topology. Assume $(X \times Y, \tau_*)$ is any space on the Cartesian product with the property that for any continuous functions $f: Z \to X$ and $g: Z \to Y$, the function $(f \times g): Z \to (X \times Y, \tau_*)$ is continuous. You want to show that the product topology $\tau_p$ is finer than the topology $\tau_*$, or in terms of functions, that the identity function $\text{id}: (X \times Y, \tau_p) \to (X \times Y, \tau_*)$ is continuous. Can you find a way to use the property of $\tau_*$ to obtain such a function?

Stefan Hamcke
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  • I fixed the notion. Thank you for pointing that out. I still don't see how to tackle the last part of the problem, but I'll try to wrap my head around your suggestion. – mattosmat Sep 05 '16 at 05:32