Homeomorphism between two spaces I am asked to show that $(X_{1}\times X_{2}\times \cdots\times X_{n-1})\times X_{n}$ is homeomorphic to $X_{1}\times X_{2}\times \cdots \times X_{n}$. My guess is that the Identity map would work but I am not quite sure. I am also wondering if I could treat the the set $(X_{1}\times X_{2}\times \cdots\times X_{n-1})\times X_{n}$ as the product of two sets $X_{1}\times X_{2}\times \cdots\times X_{n-1}$ and $X_{n}$ so that I could use the projection maps but again I am not sure exactly how to go about this. Can anyone help me?
 A: Let us denote $A = X_1\times \cdots \times X_{n-1}$ and $X = X_{1}\times \cdots\times X_{n-1}\times X_n$. The box topology $\tau_A$ on $A$ is defined by the basis of open product sets:
$$
\mathcal B(A) = \{B_1\times\cdots \times B_{n-1}:B_i \text{ is open in } X_i,1\leq i\leq n-1\}.
$$
The box topology $\tau_X$ on $X$ is defined by the basis:
$$
\mathcal B(X) = \{B_1\times\cdots\times B_{n}:B_i \text{ is open in } X_i,1\leq i\leq n\}.
$$
Let us follow Henning and put $f:A\times X_n\to X$ as
$$f((x_1,\ldots,x_{n-1}),x_n) = (x_1,\ldots,x_n)$$ 
so 
$$
f^{-1}(x_1,\ldots,x_n) = ((x_1,\ldots,x_{n-1}),x_n).
$$ 
Clearly, it is a bijection. Then we should check that $B\in\tau'$ iff $B\in \tau_X$.
Let us check it:


*

*if $B\in\tau_X$ then 
$$
f^{-1}(B) = \bigcup\limits_{\alpha}(B_{1,\alpha}\times\cdots\times B_{n-1,\alpha})\times B_{n,\alpha}\in \tau'$$ 
since $B_{1,\alpha}\times\cdots\times B_{n-1,\alpha}\in \tau_A$.

*if $B\in \tau'$ then
$$
B = \bigcup\limits_\alpha C_\alpha \times B_{n,\alpha}
$$
where $C_\alpha \in \tau(A)$. But we know the basis for the latter topology, so
$$
C_\alpha = \bigcup\limits_\beta C_{1,\alpha,\beta}\times\cdots\times C_{n-1,\alpha,\beta}
$$
where $C_{i,\alpha,\beta}$ are open in $X_i$, here $1\leq i\leq n-1$.
Finally we substitute these expressions and get
$$
f(B) = \bigcup\limits_{\alpha}B_{1,\alpha}\times\cdots\times B_{n-1,\alpha}\times B_{n,\alpha}\in \tau_X
$$
where we denote
$$
B_{i,\alpha} = \bigcup\limits_{\beta}C_{i,\alpha,\beta}\text{ - open in }X_i.
$$
Note that we also implicitly interchanged unions w.r.t. $\alpha$ and $\beta$.
