I am trying to think of manifolds without any embedding and I wondered when two manifolds are different.
For example, consider the unit circle embedded in $\mathbb{R}^2$: $x^2+y^2=1$. As far as I understand, this manifold is the same as the shifted circle $(x-1)^2+y^2=1$ (i.e. the same manifold under a different embedding). I tried to convince myself of this by calculating the transition maps of the atlas that was introduced on the Wikipedia page (i.e. projection onto the principal axes). What I get with the shifted embedding is the following:
$$ \chi_{top}:(0,2)\times(0,1] \rightarrow (-1,1) : (x,y) \mapsto x-1 \\ \chi_{top}^{-1}(a) = (a+1,\sqrt{1-a^2}) \\ \chi_{right}:(1,2]\times(-1,1) \rightarrow (-1,1) : (x,y) \mapsto y \\ \chi_{right}\left(\chi_{top}^{-1}(a)\right)=\sqrt{1-a^2} $$
So the transition map from $\chi_{top}$ to $\chi_{right}$ is exactly the same as the one for the unshifted unit circle. Hence, I would consider both manifolds the same.
Now let's consider another example: the unit sphere embedded in $\mathbb{R}^3$ and an ellipsoid that results from the unit sphere by scaling down the z-coordinate by $0.5$. I use the same kind of atlas as in the circle case. Then for the ellipsoid:
$$ \chi_{top}(x,y,z)=(x,y) \\ \chi_{top}^{-1}(a,b)=(a, b, 0.5 \sqrt{1-a^2-b^2}) \\ \chi_{right}(x,y,z)=(y,2z) \\ \chi_{right}(\chi_{top}^{-1}(a,b)) = (b,\sqrt{1-a^2,b^2}) $$ Again, the transition map from $\chi_{top}$ to $\chi_{right}$ is the same as the one for the sphere. Hence, I would believe both manifolds to be equal.
But here is the catch: Sphere and ellipsoid exhibit different Gaussian curvature. Gaussian curvature should be intrinsic and not rely on the embedding. Hence, if two manifolds were the same, they should have equal Gaussian curvatures. But this contradicts my previous observation.
So where did I go wrong?