By contraddiction, if $\phi$ would be an homeomorphism, then $[0,1)\setminus \{\frac{1}{2}\}$ is homeomorphic to
$\mathbb{S}^1\setminus \{\phi(\frac{1}{2})\}$
but the first space is not connected while the second is connected.
Another way can be to observe that $[0,1)$ is not compact while $\mathbb{S}^1$ is a compact Space because is a closed and limited subset of $\mathbb{R}^2$
On any topological $n$-manifold $M$, define an orientation of $M$ to be a function $\mu$ defined on $M$ such that for each input $x \in M$, the output $\mu(x)$ is one of the two generators of the infinite cyclic group $H_n(M,M-x)$, and the following property holds: for each embedded open $n$-ball $B \subset M$ there exists a generator $\mu_B$ of the infinite cyclic group $H_n(M,M-B)$ such that for each $x \in B$ the inclusion induced homomorphism $H_n(M,M-B) \mapsto H_n(M,M-x)$ maps $\mu_B$ to $\mu(x)$.
Now one proves
- Theorem (and definition): For every topological manifold $M$ exactly one of two possibilities holds: either $M$ has exactly two orientations, in which case we say that $M$ is orientable; or $M$ has no orientations, in which case we say that $M$ is nonorientable.
This theorem is one of the preliminary steps to the proof of Poincare Duality; see for example Hatcher's book "Algebraic Topology". In fact, proving that the relative homology groups $H_n(M,M-B)$ and $H_n(M,M-x)$ are infinite cyclic is also one of the preliminary steps.
Now to your question.
Let $M$ be a connected topological manifold.
If $M$ is nonorientable, then it makes no sense to ask whether a homeomorphism of $f$ preserves orientation, and the whole concept of "preserving orientation" is undefined for $M$.
If on the other hand $M$ is orientable then to say that a homeomorphism $f : M \to M$ is orientation preserving means that for either of the two orientations $\mu$ of $M$, and for any $x \in M$, the induced isomorphism $f : H_n(M,M-x) \to H_n(M,M-f(x))$ takes $\mu(x)$ to $\mu(f(x))$.
Best Answer
They are not homeomorphic. They are merely homotopy equivalent.
A way to see they are not homeomorphic is that they have different numbers of boundary components (three versus one).
A fancier way (using homology) is to consider the fact that a sphere with three holes can be embedded in the plane, which implies that the algebraic intersection number of any pair of closed loops is $0$ modulo $2$. But on a torus with one hole, it's easy to come up with a pair of curves that intersect in exactly one point, which means the algebraic intersection number is $1$ modulo $2$.
Addressing "they have thickness": if we are considering (as Parker does in the video) thickened surfaces, which you might formalize as products of a surface with an interval (and thus are 3-dimensional manifolds), then the thickened surfaces are indeed homeomorphic. They are both genus-2 handlebodies.
In general, if an orientable surface deformation retracts onto a wedge of $g$ circles (like $S^1\vee S^1$ for $g=2$) then their thickenings are homeomorphic to a genus-$g$ handlebody.
There is another way you can formalize a thickening that depends on the way a surface is embedded in $\mathbb{R}^3$, which is to thicken it into the ambient space (i.e., take a product with the normal bundle, rather than with a trivial $I$-bundle like above). You can drop the dependence on orientability with this notion of thickening. An "ambiently thickened" Mobius strip is homeomorphic to an "ambiently thickened" annulus, where both are homeomorphic to a genus-$1$ handlebody (a solid torus). But thickened in the first way, they are non-homeomorphic. The Mobius strip gives a non-orientable 3-manifold, but handlebodies are orientable.