About products in $\textbf{Top}$

Question

A question about the example below quoted below (Example 5.1.4 from Leinster): Which definition of the product topology does he use? Wikipedia defines the product topology to be the coarsest topology for which the projections are continuous. But in the last paragraph of the example below, Leinster refers to this as a "statement", not as a definition. Then, does he use the definition that says that the product topology on $X\times Y$ is the topology whose basis consists of all sets of the form $U\times V$ where $U$ is open in $X$ and $V$ is open in $Y$? (That's how Munkres defines the product topology on $X\times Y$.)

If so, how does the statement that $t\mapsto (x(t),y(t))$ is continuous iff $t\mapsto x(t)$ and $t\mapsto y(t)$ is continuous follow?

One direction seems to be clear: if both one-variable functions are continuous, then the preimage of every open set under $x$ and $y$ is open. It suffices to show that the preimage of a basic open set under $\phi(t)=(x(t),y(t))$ is open. Let $U\times V$ be such a set, and consider $\phi^{-1}(U\times V)=\{t:x(t)\in U, y(t)\in V\}$. This is the intersection of the sets $\{t:x(t)\in U\}$ and $\{t:y(t)\in V\}$, which are open by assumption. So $\phi$ is continuous.

Coversely, if $\phi$ is continuous, we need to show that $\{t:x(t)\in U\}$ and $\{t:y(t)\in V\}$ are both open, where $U$ and $V$ are any open sets in $X$ and $Y$ respectively. We know that $\phi^{-1}(U\times V)=\{t:x(t)\in U, y(t)\in V\}$ is open, but I don't see how this implies that the two sets we're interested in are open.

Answer 1

It follows from the universal property for mappings for initial topologies, which I showed and explained here. The map $t \to (f(t), g(t))$ has compositions with the two projections that are precisely $f$ and $g$ which are assumed to be continuous.

Answer 2

Leinster says that the product topology is designed so that

A function $f : A \to X \times Y$ is continuous if and only if the two coordinate functions $f_1 : A \to X$ and $f_2 : A \to Y$ are continuous.

This is is not a formal definition, but only a motivational introduction. I do not have access to his book, but I am sure he gives a proper definition later in his text.

Anyway, if we take the above assertion as the characteristic property of a topology on $X \times Y$, we cannot a priori be sure that such a topology exists and is unique. This requires a proof. How this works is indicated by the statement that the product topology is the smallest topology such that the projections $p_1, p_2$ are continuous. Call a topology making $p_1, p_2$ continuous an admissible topology (this is just an ad-hoc notation). Clearly the discrete topology is admissible and the intersection of all admissible topologies is admissible, thus indeed there exists a (unique) smallest admissible topology. It is easy to see that a basis of this topology is given by the products $U \times V$ of all open $U \subset X, V \subset Y$.

Hence if $f : A \to X \times Y$ is continuous, then trivially both $f_i = p_i \circ f$ are continuous. Conversely, if the $f_i$ are continuous, we have to prove that $f$ is continuous. To verify that, it suffices to show that $f^{-1}(U \times V)$ is open for all basic open $U \times V$. But $U \times V = U \times Y \cap X \times V$ and $f^{-1}(U \times Y \cap X \times V) = f^{-1}(U \times Y) \cap f^{-1}(X \times V) = f_1^{-1}(U) \cap f_2^{-1}( V) $.

Answer 3

Suppose your definition of the product topology on $X \times Y$ is that whose open sets are unions of the sets of the form $U \times V$ for $U, V$ open in $X, Y$.

Let $\phi: A \rightarrow X \times Y$ be a continuous function in this sense. Your question is why the one variable functions $A \rightarrow X \times Y \rightarrow X$ and $A \rightarrow X \times Y \rightarrow Y$ are continuous, correct?

This is obvious from the fact that each of the projections $X \times Y \rightarrow X$ and $X \times Y \rightarrow Y$ are continuous, since a composition of continuous functions remains continuous.

To see that, say, $X \times Y \rightarrow X$ is continuous, you need only observe that the preimage of an open set $U \subset X$ is $U \times Y$.