Prove that the product spaces $[0, 1] \times (0, 1]$ and $(0, 1] \times (0, 1]$ are homeomorphic. Let $[0, 1]$ and $(0, 1]$ be given in the standard topology.
Prove that the product spaces $[0, 1] \times (0, 1]$ and $(0, 1] \times (0, 1]$ are homeomorphic.
 A: You could use the homeomorphism on the unit square which consists of five convex-linear maps, each one sending a triangle on the left homeomorphically to the triangle with the same number on the right.

The restriction of this map to $[0,1]\times(0,1]$ should give you the homeomorphism you want.
A: Clearly $[0,1]\times (0,1]\approx [-1,1]\times(0,1]$ per $(x,y)\mapsto (2x-1,y)$. Via
$$\tag1 (x,y)\mapsto \begin{cases}\frac{x+iy}{\sqrt{1+(y/x)^2}}&\text{if }|x|\ge y\\
\frac{x+iy}{\sqrt{1+(x/y)^2}}&\text{if }|x|\le y
\end{cases}$$
this is homeomorphic with 
$$\tag2\{\,z\in\mathbb C\mid |z|\le1,\Im z>0\,\}$$
The same map $(1)$ shows that $(0,1]\times(0,1]$ is homeomorphic to 
$$\tag3\{\,z\in\mathbb C\mid |z|\le1,\Im z>0,\Re z>0\,\}$$
Squaring, i.e. $z\mapsto z^2$ is a homeomorphism from $(3)$ to $(2)$.
A: This is not a formal answer, just an hint. Draw the two loci you have. As you can see they are both squares with just a part of the boundary. In particular, in both cases you have "just one piece of border" (for example, in $[0,1] \times (0,1)$ you have two and this space is not homeomorphic to the ones you have). Morally you can deform one into the other (anyway this is just a rough idea I'm giving you).
