Let $X,Y,Z$ be topological spaces, and $$f:X \times Y \rightarrow Z$$be a function, it induces $$f_y:X \rightarrow Z$$ where $f_y(x)=f(x,y)$, and $$f_x:Y \rightarrow Z$$ where $f_x(y)=f(x,y)$. Obvious that if continuous $f$ implies continuous $f_x$ and $f_y$, but my problem is:
If $f_x$ and $f_y$ are continuous for all $x\in X$ and $y\in Y$, is $f$ continuous? (On metric space the answer is yes, but I don’t know it if $X,Y,Z$ are general topological spaces.)
This is false even in $\mathbb R^2$. A standard counterexample is the function $$ f(x, y)=\begin{cases} \frac{2xy}{x^2+y^2}, & (x, y)\ne (0,0), \\ 0, & (x, y)=(0,0).\end{cases}$$ Here, both $f(x, \cdot)$ and $f(\cdot, y)$ are continuous functions of $\mathbb R$ to $\mathbb R$ for all fixed values of $x$ and $y$, but $$ \lim_{t\to 0} f(t, t)=1\ne f(0,0), $$ so the function is not continuous at $(0,0)$.