I have a question related to box topology and product topology as follows:
The mapping $f: A\times A \rightarrow A$ is continuous if and only if for each $a\in A$ we have the map $f_{a} : A \rightarrow A$ in which $f_{a}(b)=f(a,b)$.
The topology defined on $A$ is product topology and the mapping is product, but first I want to see if more general form of this statements it ture or not.
It is not generally true that a function $f:S\times S\to S$ is continuous if $f_a(b)=f(a,b)$ is continuous for each $a\in A.$ Even if $f'_b(a)=f(a,b)$ is also continuous for each $b.$
Here is an example: First we need a small lemma.
LEMMA. With the usual topology on $\mathbb R,$ if $A$ is dense and if $B$ is open and not empty , then $\{x+y: x\in A\land y\in B\}=\mathbb R.$
Now let $\tau$ be the usual topology on $\mathbb R$ and let $D$ be the set of dense open subsets. Let $\tau_D=D\cup \{\phi\}.$ Then $\tau_D$ is a (rather weak) topology on $\mathbb R.$
Let $S$ be $\mathbb R$ with the $\tau_D$ topology. Let $f(a,b)=a+b$ for $a, b\in S.$ It should be obvious that $f_a(b)=a+b $ is continuous for each $a\in S.$
Now $S$ \ $\{0\}\in \tau_D.$ So if $f$ is continuous then $U=f^{-1}(S$ \ $\{0\})$ must be open in $S\times S$ . But since $U\ne \phi,$ that would imply $U\supset A\times B$ for some $A,B\in D.$ But this would imply that $0\not \in \{f(x,y):x\in A\land y\in B\},$ which contradicts the lemma.
Remark: Part of the def'n of a topological group $G$ requires that $f(a,b)=ab$ is continuous from $G\times G$ to $G$. This is stronger than requiring $f_a(y)=ay$ and $f'_b(x)=xb$ to be continuous for each $a,b,$ as the above example shows: The additive group $\mathbb R$ with the $\tau_D$ topology is not a topological group.