Identity homeomorphism in proving a topological space is a topological manifold Let $M$ be a topological space. Given $M=R^n$, it is a $n$ topological manifold because $M$ is Hausdorff, its topology contains a countable basis and $M$ is homeomrphic to $R^n.$ For the latter let $U=M, V=R^n$, then $x=\mathrm{id}$, $x:U \rightarrow V$ is a homeomorphism.
Lets take the sphere in $R^3$. By the same reasoning as above we find that the Sphere in $R^3$ is a 3 topological manifold, instead of looking for a complicated chart or homeomorphism. Maybe one could go on with other topological spaces.
My questions are:


*

*Why one almost never considers this homeomorphism/chart except for the case $M=R^n ?$

*If in the case of the sphere in $R^3$ we take that homeomorphism, we conclude that the sphere is a $3$ dim topological manifold, whereas usually it is considered as 2 dimensional manifold. Is it a mistake to consider the sphere as a $3$ dim topological manifold if one consideres the identity as a homeomorphism ?

*If the reasoning of using that (trivial) homeomorphism is correct, can you give an example of a topological manifold for which you can not take that identity homeomorphism in order to prove a topological space is a topological manifold ?
Many thanks.

Can you exhibit one open set in $\mathbb{R}^3$ that is in the domain of the identity map to $S^2 \subset \mathbb{R}^3$?  
Homeomorphisms are continuous open maps, so they map open sets from open sets and open sets to open sets.  The preimages of open balls in $S^2$ are (homeomorphic to) disks in $\mathbb{R}^3$ using the identity map.  Disks are not open in $\mathbb{R}^3$.  In the other direction, there are no open sets of $\mathbb{R}^3$ that are mapped to $S^2$ by the identity map.  So the identity map is not a homeomorphism between $\mathbb{R}^3$ and $S^2$.
