Prove that $[0,1) \times [0,1)$ is homeomorphic to $[0,1] \times [0,1)$

Question

I need to prove that:

$[0,1) \times [0,1)$ is homeomorphic to $[0,1] \times [0,1)$

I am not confident in writing homeomorphisms, but I picture the situation as:

I noticed that the each side of the half open (on the left) and U-shaped (on the right) squares is homeomorphic to another, then the interior of these squares are the same so one can use the identity map in there.

However, I could not manage to write a specific homeomorphism $f$ for the mentioned sides.

Answer 1

Here is my Intuitive way to see the "mapping" , where I am giving the high-level operations which are required.

You should be able to make the necessary functional relationship with that. Let me know if you encounter issue with that.

Convert every Blue Solid line $y=x+a$ ( $-1.0<a<+1.0$ ) to half-way to axis along the line : We will get the Blue Dotted line.
We have got the triangle.

Convert every Grey line $y=x+a$ ($-0.5<a<+0.5$) to half-way to diagonal. The other $a$ value do not change.
We have got the shaded area.

Convert every Diagonal Purple line $y=-x+a$ ($-0.5<a<+1.0$) to half-way towards the Diagonal $y=x$ to get the shaded area.

Rotate by $45^\circ$ to get this :

That can easily map to the required area.

Combine all the invertibile maps to get the required homeomorphism.
Do let me know if you encounter Issues with that.

Over-all view :
We are shrinking various lines either towards axis or towards middle. We will not shrink certain lines which are already where we want them.
We will then rotate to align with required area.
That gives easy way to map two rectangles with the top line missing.
Putting all these maps together , we get the single source-to-destination invertiable map.

Answer 2

My approach would be to prove two lemmas:

• Every pair of triangles in the plane are homeomorphic (use a linear transformation).

• If two spaces may be covered by finitely-many homeomorphic closed subspaces, such that the homeomorphisms coincide where they overlap, then the union of these homeomorphisms is itself a homeomorphism.

Then I'd draw a picture like this:

Then we have a homeomorphism that sends the blue path to the blue path; restrict to the rest of the square to obtain the desired map.