Let $X$ be a set and $\{\tau_i\}_{i\in I}$ a family of topologies on $X$. If the set map $f:X\times X\to X$ is continuous w.r.t. $\tau_i$ ($X\times X$ given product topology) for all $i$,then will $f$ be continuous w.r.t. $\tau:=\cap_i\tau_i$?
So I take an open set $U$ w.r.t. $\tau$, $f^{-1}(U)$ contains an basic open set $V_i\times W_i$ wrt $\tau_i\times\tau_i$ and now a subset inside these $V_i,W_i$ which are open in all $i$. I am like unable to continue after this. I don't think its even true tbh. I also thought of using nets but if a net converges in each topology will it converge w.r.t intersection?
Here is a counterexample. Let $X=\mathbb{N}\cup\{a,b,c,d\}$. Let $\tau_0$ be the topology generated by the sets $\{n\}$ for each $n\in\mathbb{N}$, $\{a\}$, $\{c\}$, and $\{d\}\cup S$ where $S$ can be any cofinite subset of $\mathbb{N}$. Let $\tau_1$ be defined the same way except with the roles of $c$ and $d$ reversed. Let $$A=\{(c,d)\}\cup(\{c\}\times\mathbb{N})\cup(\mathbb{N}\times\{d\})\cup\{(m,n)\in\mathbb{N}\times\mathbb{N}:m\neq n\}$$ and let $f:X\times X\to X$ be the function that maps $A$ to $a$ and $X\setminus A$ to $b$. I claim first that $f$ is continuous with respect to both $\tau_0$ and $\tau_1$, which just amounts to saying that $A$ is open in both product topologies. To see $A$ is open with respect to $\tau_0$, the only nontrivial part is to see that it contains an open neighborhood of $(c,d)$ and of $(n,d)$ for each $n\in\mathbb{N}$. But this is true since it contains $\{c\}\times(\{d\}\cup\mathbb{N})$ and $\{n\}\times(\{d\}\cup\mathbb{N}\setminus\{n\})$. The proof that $A$ is open with respect to $\tau_1$ is similar.
However, $f$ is not continuous with respect to $\tau_0\cap\tau_1$, since $A$ is not open with respect to $\tau_0\cap\tau_1$. Indeed, any neighborhood of $\{(c,d)\}$ with respect to $\tau_0\cap\tau_1$ must contain a set of the form $(\{c\}\cup S)\times(\{d\}\cup T)$ where $S$ and $T$ are both cofinite in $\mathbb{N}$. Then $S\cap T$ is nonempty, so $(\{c\}\cup S)\times(\{d\}\cup T)$ is not contained in $A$ since it contains points of the form $(n,n)$ for $n\in\mathbb{N}$.