Does a differential manifold implies existence of unique tangent space at every point? I like differential geometry and I want to know if a differentiable manifold implies unique tangent space at every point. I have searched but the definition I have found of differential manifold is that differentiable manifold is a topological manifold in which we construct an atlas that covers the entire manifold and the transition functions in the intersection of the coordinate systems of the atlas are differentiable. But this doesn't say anything about tangent spaces. Whatever I have read about differential geometry, tangent spaces are defined always after the definition of differentiable manifold, so it seems tangent spaces are not need to define what a differentiable manifold is and for a topological manifold to be a differentiable manifold is only need that we construct an atlas with the characteristics I said before. But then if you have for example, a 1 dimensional topological manifold that is a real line with one point that is not smooth, because it has the form of a peak, there is no unique tangent space at this point, although there is in the other points. Then we construct an atlas in which the transition functions are differentiable at this point just  because we can define the coordinate systems in a way that the transition functions from one coordinate to another  is differentiable. Although the manifold has no unique tangent space in this point. I always believed that differentiability at manifold level implies unique tangent space and that the manifold doesn't care about coordinate system differentiability. But the definition of differentiable manifold is based on differentiability in coordinate systems. So if we can define an atlas that is smooth but the real manifold is not, Therefore, differentiable manifold doesn't implies unique tangent space. Am I missing something?
Sorry for my bad expression but english is not my mother tongue and I tried to express the better I can.
 A: $\newcommand{\Reals}{\mathbf{R}}$In the context of your question, a smooth $n$-manifold $M$ is a topological space equipped with additional intrinsic structure (locally homeomorphic to the Cartesian space $\Reals^{n}$, smooth overlap maps).
Each tangent space of $M$ is defined in terms of (and uniquely characterized by) this intrinsic structure. In that sense, $M$ has a unique tangent space at each point.
A smooth manifold does not come equipped with an embedding into a Cartesian space. In fancier words, the underlying set of points of a smooth manifold need not be a subset of $\Reals^{N}$ for some positive integer $N$.
If $M$ is a $n$-smooth manifold, and if $\phi:M \to \Reals^{N}$ is a smooth embedding, then the image $\phi(M) \subset \Reals^{N}$ is locally homeomorphic to $\Reals^{n}$, and these homeomorphisms may be chosen so that the overlap maps are smooth. In this setting, the intrinsic tangent space of $M$ at a point $p$ corresponds naturally with a certain affine subspace in $\Reals^{N}$ passing through $\phi(p)$. This is the intuitive picture underlying your question. Note carefully, though, that you are no longer considering a smooth manifold alone, but a smooth manifold together with a smooth embedding in $\Reals^{N}$.
Consider a subset $M \subset \Reals^{N}$ that is locally homeomorphic to $\Reals^{n}$, such as the graph $y = |x|$ in $\Reals^{2}$. It may be true that $M$ can be given the structure of a smooth manifold (by fixing, at each point $p \in M$, one homeomorphism from some neighborhood of $p$ to $\Reals^{n}$, then taking all "smoothly compatible" homeomorphisms), but even if you do so, it does not follow that the inclusion map $i:M \to \Reals^{N}$ is smooth. (!) That is, the intrinsic tangent spaces of $M$ need not correspond with affine subspaces of $\Reals^{N}$. This resolves the apparent paradox implicit in your question.
A: Fix two different parameterizations $p: U \longrightarrow M,q:V \longrightarrow M$ around a point $x$ on a smooth manifold $M$. The corresponding points $u_1 \in U, v_1 \in V$ lie in Euclidean spaces where tangents exist naturally. Pick a tangent vector to $u_1$, that is a vector $\vec{u_1}$ with its base at $u_1$. The transition map $U \longrightarrow V $ takes $u_1$ to $v_1$ and the vector $\vec{u_1}$ to a new vector $\vec{v_1}$ tangent to $V$ at $v_1$. Since transition map is a local diffeomorphism, its derivative is 1-to-1 and this gives a pairing of tangent vectors on $U$ and $V$.
The tangent to $M$ at $x$ can be thought of as tangent to $U$ at $u_1$ for some parameterization of $M$ around $x$. To have  well-defined notion we identify such $\vec{u_1}$ and $\vec{v_1}$ pairs under transition of parameterizations. In other words, one tangent vector to $M$ at $x$ is the classof all tangent vectors  to localizations of $M$ where we identify two vectors in two parameterizations iff they map to one another under transition of maps.
A: The differentiable structure on the manifold does determine (uniquely) its tangent space.
As for familiar broken shapes, it depends how you look at them. If you only see them as topological subsets of Euclidean spaces, that is all you borrow from $R^n$ is the topology, then you can (yourself) ENDOW them with a differentiable atlas and structure
For instance, a rectangle, when we see it as a topological space is no different than a circle. Then we can transfer the differentiable structure of $S^1$ onto the rectangle via their homeomorphism. However, the circle is by default a differentiable submanifold of $R^2$, that is, with $R^2$'s atlas restricted to the circle, we do get an atlas on $S^1$. The problem with the rectangle is we may equip it with a differentiable structure, but it won't be the restriction of the plane's atlas to the rectangle. This is a relief! Since our intuition of rectangle being non-smooth is respected by our math definitions.
Addition: You may be surprised to hear that there exists topological manifolds that do not admit any smooth structures! That is among so many spaces that a fixed space is homeomorphic to, non is a smooth manifold!
