Definition of vector fields on submanifolds There is a definition in my lecture which I don't really understand:
Let $\overline{M}$ be a manifold and $M$ a submanifold. Now I'm "quoting" (+translating) the lecture notes:
Each vector field $X \in \mathfrak{X}(\overline{M})$ defines a vector field $X_{|M}$ along the inclusion $M \hookrightarrow \overline{M}$. Define $\overline{\mathfrak{X}}(M)$ as the $C^{\infty}(M)$ module of vector fields along the inclusion $M \hookrightarrow \overline{M}$. Also, it holds $\mathfrak{X}(M) \subset \overline{\mathfrak{X}}(M)$.
My interpretation: $X \in \mathfrak{X}(\overline{M})$ means that $X$ is a function $\overline{M} \rightarrow T\overline{M}$. Now $X_{|M}$ is then a function $M \rightarrow \bigcup\limits_{p \in M} T_p \overline{M}$ right? A vector field $X \in  \mathfrak{X}(M) $ is a function $M \rightarrow TM$. Thus $TM \subset \bigcup\limits_{p \in M} T_p \overline{M}$, right? Now that's something I don't understand. Shouldn't it be $TM =\bigcup\limits_{p \in M} T_p \overline{M}$?
I'm not quite sure, if I understand what $\bigcup\limits_{p \in M} T_p \overline{M}$ means.
Sorry for my sometimes confusing explanations but maybe still somebody can explain this to me..
 A: Could be too late, but I'm thinking about it.
For the inclusion case, we can see $X|_M$ as a vector field of $\bar{M}$ restricted to $M$. Note that $X|_M(p)$, in general, does not belong to $T_pM$, but always to $T_p\bar{M}$. So $$TM=\bigcup_{p\in M} T_pM \subsetneq \bigcup_{p \in M}T_p\bar{M}\subseteq \bigcup_{p\in\bar{M}}T_p\bar{M}=T\bar{M}.$$
Think about $\mathbb{S}^2\subset\mathbb{R}^3$, and $X=(x,y,z)\in\mathfrak{X}(\mathbb{R}^3)$. In that case, $X|_{\mathbb{S}^2}$ is a vector field along the inclusion but the vectors of this field are normal to the sphere, so they aren't tangent and $X|_{\mathbb{S}^2}\notin \mathfrak{X}(\mathbb{S}^2)$. For a geometric intuition of what $\bigcup_{p \in \mathbb{S}^2} T_p\mathbb{R}^3$ means, think it as the space of $$\{ (p,v) \in \mathbb{R}^3\times\mathbb{R}^3:p \in \mathbb{S}^2 \text{ and } v \in \mathbb{R}^3\}.$$
So, you kinda have an $\mathbb{R}^3$ attached to each point on the sphere, and not an $\mathbb{R}^2$ (tangent plane to be more precise) attached to each point, as $T\mathbb{S}^2$ suggests.
For the general case $f:\bar{M}\rightarrow \bar{N}$, a priori, we dont need to consider the restriction. In fact, if $Y \in \mathfrak{X}(\bar{N})$, it's not quite that $Y\circ f$ is defined on the image of $f$. The vector field $Y\circ f:\bar{M}\rightarrow T\bar{N}$ along $f$ is a function defined on $\bar{M}$ and not a vector field in $\bar{N}$ restricted to $f(\bar{M})$.
