I'm wondering how the manifold topology on the tangent bundle $ TM $ yields a notion of continuity between tangent vectors as well, and not merely between base points. Let $X: M \rightarrow TM$ be a function such that $X(m) \in T_mM$, let's check when it is continuous.
As the sets $ TU $ are open and form a basis of the topology of $ TM $ (for admissible charts $ (U, \phi ) $ of $ M$), one only needs to check that $X_{}^{-1}(TU) $ are open in $ M $. This is true, as $X_{}^{-1}(TU) = \{ m \in M | (m, \text{ some tangent vector at } m) = X(m) \in TU \} = U$. Here I have used that $ TU $, for any point $ u \in U $, contains all tangent vectors of $ T_u M$, so that what is important in $X(m) $ is just the base point.
Therefore I would say that the "vector part" of a function into $ TM $ is irrelevant for continuity, as long as the "base point" part is continuous on $M$. This would however make the whole notion of smooth vector field pointless, so there must be a mistake somewhere in my reasoning. Thanks in advance!
My definition of topology on a manifold is the one generated by the charts in the maximal atlas.
In response to Arctic Char
- The functions $\phi^*: TU \rightarrow \phi(U) \times \mathbb{R}^n$ are bijective, for $\phi, U$ a chart in the maximal atlas of $M$. The sets $TU$ cover $TM$ and two such charts are compatible, hence $\phi^*, TU$ form an atlas on the (for now) set $TM$.
- Thus $TM$ receives a differentiable structure from this atlas, and also a topology, given by the sets $W \subseteq TM$ such that for every $v \in W$, one can find a chart in the maximal atlas of $TM$ whose domain is entirely contained in $W$. Call this topology $\tau$.
- As you said, here an apparently different topology is constructed. The topology in exam is $\tau':=\{W\subseteq TM: \phi^*(W\cap TU)\ \textrm{is open in}\ \mathbb R^n\times \mathbb R^n\ \forall\ (U, \phi) \text{ charts of }M\}$. I want to prove $\tau=\tau'$. Let's check the two inclusions.
- Suppose $W$ is open in the topology $\tau'$. Thus, $\phi^*(W\cap TU)\ \textrm{is open in}\ \phi(U) \times \mathbb R^n$, and since $\phi^*$ is a diffeomorphism, $W\cap TU$ is open in my topology $\tau$, so that also $W$ is open, as a union of open sets.
- If on the other hand $W$ is open according to $\tau$, as $\phi^*$ is a diffeomorphism and $TU$ is open, also $\phi^*(W\cap TU)\ \textrm{is open in}\ \phi(U) \times \mathbb R^n$.
You have to be careful, the sets $TU$ are open in $TM$, but they do not form a basis on the topology of $TM$: the open sets in $TM$ are defined (see here for example) so that $$ \phi_* : TU \to T (\phi (U)) \cong \phi (U) \times \mathbb R^n$$ is a homeomorphism. Thus e.g. sets of the form
$$ \phi_*^{-1} (U \times V),$$
where $V$ is open in $\mathbb R^n$ is also open in $TM$. In local coordinates, $ \phi_* X \phi^{-1} : \phi(U) \to \phi(U) \times \mathbb R^n$ is graph. That is,
$$ \phi_* X \phi^{-1} (v) = (v ,F(v))$$ for some $F : \phi (U) \to \mathbb R^n$. Continuity of $ \phi_* X \phi^{-1} $ is equivalent to the continuity of $F$.