Product topology on $X \times Y$ the smallest topology when $f(x, y) = x$ and $g(x, y) = y$ are continuous functions?

Question

$X$ and $Y$ are topological spaces and let $f : X \times Y \to X$ and $g : X \times Y \to Y$ be maps such that $f(x, y) = x$ and $g(x, y) = y \ \forall (x, y) \in X \times Y$ . Show that the product topology on $X \times Y$ is the smallest topology on $X \times Y$ for which both the $f$ and $g$ are continuous.

I think I understand what is going on here. Here is how I see it.

As $f$ is continuous, take an open set $U \subset X$ and it's inverse image $(U, Y)$ will be open in $X \times Y$ for any topology on $X \times Y$.

Similarly, as $g$ is continuous, take an open set $V \subset Y$ and it's inverse image $(X, V)$ will be open in $X \times Y$ for any topology on $X \times Y$.

Hence $(U, Y) \bigcap (X, V) = (U, V)$ which will be open in $X \times Y$ for any topology on $X \times Y$.

Now the set $\{(U, V): U \in X, V \in Y\}$ is the product topology hence we are saying that the product topology will be a subset of any other topology on $X \times Y$ when $f$ and $g$ are the continuous maps as defined above.

Is my understanding correct? If so is my proof clear or is it messy?

Answer 1

The product topology on $X \times Y$ is defined by giving a basis for the topology, namely the set ${\mathcal B} := \{ U \times V \mid U \subseteq X \text{ open}, V \subseteq Y \text{ open} \}$.

(As mentioned in the comments, one could also define the product topology to be the coarsest topology that makes the projections continuous. From the remainder of the question and the comments by the OP, however, it becomes clear that the OP defines the product topology via the basis ${\mathcal B}$.)

Now suppose we have an alternative topology on $X \times Y$; let's call this topological space $Z$ (so it has the same underlying set as $X \times Y$). We assume that the projections $f \colon Z \to X$ and $g \colon Z \to Y$ are continuous and we want to show that every open subset $X \times Y$ is also an open subset of $Z$.

It is sufficient to show this for the open subsets in ${\mathcal B}$, so take an open subset $U$ of $X$ and an open subset $V$ of $Y$. We must argue that $U \times V$ is open in $Z$.

Your argument now essentially works.

Because $U$ is open in $X$, its preimage $f^{-1}(U) = U \times Y$ is open in $Z$. Similarly, because $V$ is open in $Y$, its preimage $g^{-1}(V) = X \times V$ is open in $Z$. Now $U \times V = (U \times Y) \cap (X \times V)$ is also open in $Z$, as required.

(As an aside, the mistakes in your argument are not in your understanding of the argument, but in its presentation: you're mixing $\cap$ and $\times$ at occasion and you didn't mention that the set I call ${\mathcal B}$ above is only a basis for the product topology.)

Alternatively, you could use the universal property of the product. Since $f \colon Z \to X$ and $g \colon Z \to Y$ are continuous, there must be a unique continuous map $u \colon Z \to X \times Y$ such that $\pi_1 \circ u = f$ and $\pi_2 \circ u = g$.

Considered as maps of sets, the only candidate for $u$ is the identity $i \colon Z \to X \times Y$ and therefore the identity map $i \colon Z \to X \times Y$ is continuous. So, given an open set $U \subseteq X \times Y$, its preimage $U = i^{-1}(U) \subseteq Z$ is open.

Answer 2

Your proof seems to be essentially correct, but I find your notation quite weird. You seem to read $(U,V)$ as $U \times V$, at least sometimes, so I cannot really say if your proof is formally correct. Here's how I'd write that proof

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Let $T$ be the box topology on $X\times Y$, i.e. the topology generated by the base $$ \left\{U\times V \,:\, \text{$U$ open subset of $X$, $V$ open subset of $Y$}\right\} \text{.} $$ (This is what you call the product topology on $X \times Y$. Note, however, that the box and product topologies are the same only for the product of finitely many spaces!)

If $U \subset X$ is open it follows that $f^{-1}(U) = U \times Y$ is open (i.e. $\in T$), and similarly if $V \subset Y$ is open so is $g^{-1}(V) = X  \times V$, hence $f$ and $g$ are continuous. Thus what remains to be shown is that $T$ is the smallest topology on $X \times Y$ with that property.

For $f$ and $g$ to be continuous, a topology $T'$ on $X \times Y$ must at the very least satisfy $$ T' \supset \left\{ f^{-1}(U) \,:\, \text{$U \subset X$ open}\right\} \cup \left\{ g^{-1}(V) \,:\, \text{$V \subset Y$ open}\right\}, $$ in other words the preimage under $f$ and $g$ of open sets in $X$ respectively $Y$ must be open. You can also write that as $$ T' \supset \left\{U\times Y \,:\, \text{$U$ open subset of $X$}\right\} \cup \left\{X\times V \,:\, \text{$V$ open subset of $Y$}\right\} \text{.} $$ But since the intersection of finitely many open sets is open, and since $U \times Y \cap X \times V = U \times V$, that implies that $$ T' \supset \left\{U\times V \,:\, \text{$U$ open subset of $X$, $V$ open subset of $Y$}\right\} \text{.} $$ So for every topology $T'$, $f$ and $g$ being continuous implies that $T' \supset T$, and $T$ is thus indeed the smallest such topology.