Is homeomorphism equivalent to uniqueness of tangent plane for each point? I have two differential geometry books, in the first one, they give the conditions for the surface they are going to use:

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*$X(u,x)=(x(u,v),y(u,v),z(u,v))$ is differentiable, that is: $x,y,z$ have continuous partial derivatives of all orders.

*$X$ is homeomorphism.

*The differential is injective.

In the second book, we have:

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*$X(u,x)=(x(u,v),y(u,v),z(u,v))$ is differentiable of class $C^{\infty}.$

*The differential is injective.

Although the second book doesn't have the condition for homeomorphism, it reads in another part of the text:

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*We are going to restrict our study to surfaces that such that in each point, have one tangent plane.

Question: Does homeomorphism $\Leftrightarrow$ each point admits one tangent plane? Intuitively, it seems to be so because if $X$ is homeomorphism, then the surface do not have self-intersections and such and hence, each point have one tangent plant. Now, if each point in the surface have one tangent plane, then we can't construct two different tangent planes, each one approaching from a different direction and hence, there won't be self-intersections and such.
 A: The first book requires that $\bar X : D \stackrel{X}{\to} X(D) =S $ is a homeomorphism (where the domain $D \subset \mathbb R^2$ is some open set). In my opinion this is the standard definition.
It is not true that

$\bar X$ is a homeomorphism $\iff$ Each point $\eta \in S$ admits one [exactly one!] tangent plane.

If you do not require that $\bar X$ is a homeomorphism, then $X$ is not necessarily an embedding, but only an immersion which is charcterized by the injectivity of the differential. This means that $\bar X$ is a local diffeomorphism. That is, each point $\xi \in D$ has an open neigborhood $U \subset D$ such that $V = \bar X(U)$ is open in $S$ and $\bar X$ maps $U$ diffeomorphically onto $V$. Then $\eta = \bar X(\xi)$ has a unique tangent plane in the surface piece $V$. Let us call it the tangent plane at $\eta$ associated to the preimage $\xi \in \bar X^{-1}(\eta)$.
A given $\eta \in S$ may have more than one preimage in $D$. If $\xi_1, \xi_2$ are two distinct preimages, then $\eta$ has  unique tangent planes in the surface pieces $V_i = \bar X(U_i)$, but these tangent planes may not be the same.
If $\bar X$ is required to be a homeomorphism, then each $\eta \in S$ has exactly one preimage $\xi \in D$ and therefore exactly one tangent plane.
Let us now consider the map
$$X : \mathbb R^2 \to \mathbb R^3, X(u,v) = (\cos u, \sin u, v).$$
This is an immersion, but no homeomorphism because $X(u,v) = X(u +2k\pi, v)$ for all $k \in \mathbb Z$. Nevertheless each $\eta \in S= X(\mathbb R^2)$ has a unique tangent plane. In fact, $\eta =(x,y,z)$ has infinitely many preimages $\xi_k = (u_k, z)$, where $u_k = u_0 + 2k\pi$. The sets $U_k = (u_k -\pi, u_k+\pi) \times \mathbb R$ are open neighborhoods of $\xi_k$ which have the same image under $X$, and this results in identical tangent planes associated to all preimages of $\eta$.
