Suppose we have spaces $X, Y$ and a topology $\tau$ on $X \times Y$. Can we find a topology $\tau_X$ on $X$, $\tau_Y$ on $Y$ such that $\tau_X \times \tau_Y = \tau$ (the product is referring to the product topology).
I think not. What if $\tau$ is a separable complete metric space?
I could also get away with somemthing weaker, like $\tau \subseteq \tau_X \times \tau_Y$ provided $\tau_X$ and $\tau_Y$ have sufficiently neat property (i.e, they're not discrete topologies or something).
The answer is no even when $X=Y=\Bbb R$, and $\tau$ is a separable metric topology on $X\times Y$.
Let $\varphi:\Bbb R^2\to\Bbb R$ be a bijection, let $\tau_e$ be the Euclidean topology on $\Bbb R$, and let $\tau=\{\varphi^{-1}[U]:U\in\tau_e\}$; let $X$ denote $\Bbb R^2$ equipped with the topology $\tau$. Then $X$ is homeomorphic to $\langle\Bbb R,\tau_e\rangle$, so $X$ is path connected, and $X\setminus\{p\}$ is not path connected for any $p\in X$.
Let $\tau_0$ and $\tau_1$ be topologies on $\Bbb R$. If $\tau$ is the the product topology on $\Bbb R^2$ generated by $\tau_0$ and $\tau_1$, then $\langle\Bbb R,\tau_i\rangle$ must be path connected for $i\in\{0,1\}$. But it’s easily seen that if $Y$ and $Z$ are path connected, and $p\in Y\times Z$, then $(Y\times Z)\setminus\{p\}$ is path connected. (E.g., you can use the idea that I used in this answer, or the one that I used in this answer.) Thus, the product topology on $\Bbb R^2$ generated by $\tau_0$ and $\tau_1$ cannot be $\tau$.
For a rather different argument, notice that if the product topology were $\tau$, then $\langle\Bbb R,\tau_1\rangle$ would have to be connected. But then $\big\{\{x\}\times\Bbb R:x\in\Bbb R\big\}$ would be a partition of $X$ into uncountably many non-trivial connected sets, which is impossible: $X$ is homeomorphic to $\Bbb R$, every non-trivial connected subset of $\Bbb R$ contains a non-empty open interval, and $\Bbb R$ does not contain any uncountable family of pairwise disjoint, non-empty open intervals.
One notable fact about the product topology is that it is the weakest topology that makes both projection maps continuous. Also note, if X and Y have topologies A and B so that the projection maps from $X\times Y$ are continuous, then $A\times B$ is contained inside the topology on $X\times Y$, C. So if C is to be written as a product of A and B, then A and B should be chosen to be the strongest topologies that make the projection maps continuous.
As a counterexample, let X and Y both be the real line. Endow $X\times Y$ with the topology generated by the standard topology and open intervals on the diagonal. This topology is strictly larger than the standard topology however if we endow both X and Y with the strongest topologies that make the projections continuous they would have the standard topologies on the real line.
Notation: $P(Z)$ (the power-set of $Z$) is the set of all subsets of a set $Z.$
If $\tau=\tau_x\times\tau_Y$ then $\tau \subseteq P(X)\times P(Y),$ which is generally not the case. Example: Suppose $X=Y=\{1,2\}$ and $s=\{(1, 2),(2,1)\}\in \tau.$ If $A\in P( X)$ and $B\in P (Y)$ and $s\subseteq A\times B$ then $(1,2)\in s\implies 1\in A$ and $(2,1)\in s\implies 2\in A,$ so $A=X.$ Similarly $B=Y.$ So $A\times B=X\times Y\ne s.$
Notation: $f[S]=\{f(s):s\in S\}$ when $f$ is a function and $S\subseteq dom(f).$
Given a topology $\tau$ on $X\times Y$: For $(x,y)$ let $p_X(x,y)=x$ and $p_Y(x,y)=y$... ($p_X$ and $p_y$ are the projections of $X\times Y$ to its $X$ & $Y$ co-ordinates.)... Let $\tau_X=\{p_X[s]:s\in \tau \}$ and $\tau_Y=\{p_Y[s]: s\in \tau\}.$ Then $\tau_X, \, \tau_Y$ are the strongest topologies on $X,Y$ such that $p_X,\,p_Y$ are continuous. And $\tau_X\times\tau_Y$ is a base (basis) for $\tau.$
