Recycling an old (ca. 1998) sci.math post:
": Anyone know an example of two topological spaces X$X$ and Y$Y$ : withwith continuous bijections f:X-->Y$f:X\to Y$ and g:Y-->X$g:Y\to X$ such that : f$f$ and g$g$ are not homeomorphisms?
Let X = Y = Z x {0,1}$X = Y = Z \times \{0,1\}$ as sets, where Z$Z$ is the set of integers. We declare that the following subsets of X$X$ are open for each n>0$n>0$. {(-n,0)} {(-n,1)} {(0,0)} {(0,0),(0,1)} {(n,0),(n,1)}$$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a topology on X$X$.
We declare that the following subsets of Y$Y$ are open for each n>0$n>0$. {(-n,0)} {(-n,1)} {(0,0),(0,1)} {(n,0),(n,1)}$$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a toplogy on Y$Y$.
Define f:X-->Y$f:X\to Y$ and g:Y-->X$g:Y\to X$ by f((n,i))=(n,i)$f((n,i))=(n,i)$ and g((n,i))=(n+1,i).$g((n,i))=(n+1,i).$ Then f$f$ and g$g$ are continuous bijections, but X$X$ and Y$Y$ are not homeomorphic.
This example is due to G. Paseman.
David Radcliffe "
More generally, take a space X with three successively finer topologies T, T' and T''. Form two spaces which have underlying set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' .... and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer topology in one sequence to a rougher topology in the other. You can make them bijective, and show that they are obviously non-homeomorphic for a judicious choice of X, T, T', and T''.
Gerhard "Ask Me About System Design" Paseman, 2010.07.05
