Is $f$ a homeomorphism? 
Problem. Let $f_i:X_i\to Y_i$ be homeomorphisms from $X_i$ to $Y_i$ where $X_i$'s and $Y_i$'s are arbitrary topological spaces. Let, $X=X_1\times X_2\times\cdots\times X_n$ and $Y=Y_1\times Y_2\times\cdots\times Y_n$. Prove or disprove the function $f:X\to Y$ defined by, $$f(x_{1},x_2,\ldots,x_n)=(f_1(x_1),f_2(x_2),\ldots,f_n(x_n))$$ is a homeomorphism. Here $X$ and $Y$ are product spaces.  

So far I have been able to show only that $f$ is a bijection. If I assume that the result is true then to show that $f$ is continuous, I need to take an open set of $V$ of $Y$ and show that its pullback is also open in $X$. So far, I haven't been able to do this. Any hint regarding this will be appreciated.
If the result is not true in general then what conditions on $X$ (or on $X_i$'s)     will ensure that the result holds?
 A: A function $f:A \rightarrow \Pi_i Y$ is continuous iff every coordinate is. We have that every coordinate is $f_i\circ \pi_i$. Since composition of continuous functions is continuous, $f_i \circ \pi_i$ is continuous. Then, we have that $f$ is continuous.
To show that $f^{-1}$ is continuous is analogous.
A: Note that it is enough to prove that $f^{-1}(V)$ is open where $V$ is a basic open in the product topology of $Y$. So, if $V=\prod_{i=1}^n V_i$ if one of such basic opens, then $$f^{-1}(V)=\{(x_1,\ldots,x_n): f(x_1,\ldots,x_n)\in V\}=\{(x_1,\ldots,x_n):f(x_i)\in V_i\}=\prod_{i=1}^n f^{-1}_i(V_i)$$ is an open in $X$ because each $f_i^{-1}(V_i)$ is an open in $X_i$.
A: It is a homeomorphism if you endow $X$ and $Y$ with the product topology which makes every projection $p_i: X \to X_i$ continous. If you are dealing wiht a finite number of spaces, the product topology coincides with the so-called box topology. 
So your argument could be '$f$ is continuous because it is continous coordinatewise'. The same for its inverse. 
A: Let $V$ be an open subset of $Y$, and let $\eta=(\eta_1,\cdots,\eta_n)\in V$, coming from $\xi\in X$. There’s an $n$-tuple of open subsets $V_i\subset Y_i$, with each $\eta_i\in V_i$, such that $\prod_iV_i\subset V$ — that’s how the product topology is defined. Now $f_i^{-1}V_i=U_i$, open in $X_i$ by continuity of $f_i$, and $\prod_iU_i$ is an open neighborhood of $\xi$ contained in $f^{-1}V$. I think that does it.
