$X$ 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?
Best Answer
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
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.