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Let $M$ be a manifold of dimension $m$ and class $C^k (k\ge 0)$. Given a point $p\in M$, let's define the tangent vector space $T_pM\subset \mathbb R^n$ at $p$.

Given a parametrization $\varphi:V_0\to V$ of class $C^k$ in a neighborhood $V$ of $p$ with $\varphi(a)=p$. The tangent vector space $T_pM$ is defined as the image of linear transformation $\varphi'(a)$, i.e, $T_pM=\varphi'(a)\cdot\mathbb R^m$.

I think $p\in T_pM$ because of some pictures I have seen and because of the name "tangent vector space at $p$" is very suggestive that $p\in T_pM$. I'm sorry if my question is too silly, I'm a beginner in this subject. I would like to know why $p\in T_pM$.

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    I'm confused; $p$ is certainly not in $T_p M$. What makes you think so? Also, what is meant by $f'(a)\cdot\Bbb R^m$? – 2017-02-06
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    @AlexMathers $f'(a)\cdot\mathbb R^m$ is the image of the derivative $f'(a)$ (remember in higher degrees the derivative is a linear map instead of a number) – 2017-02-06
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    @AlexMathers ha I understood your question, see my edit please – 2017-02-06
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    @AlexMathers in fact I meant $\varphi'(a)\cdot\mathbb R^m$ – 2017-02-06
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    @AlexMathers I think $p\in T_pM$ because of some pictures I have seen and because of the name "tangent vector space at $p$" is very suggestive that $p\in T_pM$ – 2017-02-06
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    $T_pM$ is a vector space. Its elements are tangent vectors. The point $p$ belongs to the manifold, not to the vector space. We refer to "tangent space at $p$" intending to say that on this point we can make a linearization of $M$. It cannot be global unless $M$ is $\mathbb{R}^n$ itself. The point is that in any point is possible to make this under certain good definitions of manifolds. – 2017-02-06

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