Give the point set with a corner at the orgin $$M:=\{(x,y)\in \mathbb{R}^2 : y=|x|, x\in (-1,1)\}$$ with the subspace topology of $\mathbb{R}^2$, $(M,\mathcal{T})$ forms a Hausdorff 1-dim locally Euclidean space. Next if we can define a differential structure $\mathfrak{F}$, then $(M, \mathcal{T}, \mathfrak{F})$ will give us a smooth manifold.
Note the projection map $\pi$: $(x,y)\mapsto x$ is a global coordinate map of $M$, it is easy to check that it is a homeomorphism. With $\mathfrak{F}_0 = \{(M, \pi)\}$ we just define $\mathfrak{F}$ to be the maximal collection such that $$\mathfrak{F} = \{(U, \phi) : \phi\circ \pi^{-1} \text{ and } \pi \circ \phi^{-1} \text{ are } C^\infty\}.$$
From the above definition, it is indeed a smooth manifold. Now if this is a manifold, when we talk about tangent spaces, how can we connect the tangent space at $(0,0)$ to the usual tangent line conecpt in calculus.
