Q:

Topology of the tangent bundle

Let $M$ be a smooth manifold, and let $\pi:TM\to M$ be its tangent bundle. We define the topology of $TM$ by declaring a subset $V$ of $TM$ to be open if and only if $\psi_\phi(V\cap\pi^{-1}(U))$ is open for each chart $(U,\phi)$ for $M$, where $\psi_\phi:\pi^{-1}(U)\to\phi(U)\times\mathbb{R}^n$ is the bijection sending $v_p\in T_pM$ (with $p\in U$) to $(\phi(p),\lambda^1,\ldots,\lambda^n)$, where the $\lambda^i$ are the coefficients of $v_p$ w.r.t to the coordinate basis $\left\{\frac{\partial}{\partial x_1}(p),\ldots,\frac{\partial}{\partial x_n}(p)\right\}$.
My teacher wrote as a comment that this topology would make the zero section become open, which is not the case, so that this was not correct. I had phrased what I state above in a somewhat ambiguous way and I think that that is why he wrote this, so I want to clarify this: this topology is the right one right? And it does not make the zero section open, since if we let $U_1$ be an open subset of $M$ and $\sigma:M\to TM$ the zero section, then for any chart $(U,\phi)$ for $M$ we have that $\psi_\phi(\sigma(U_1)\cap\pi^{-1}(U))=\phi(U_1\cap U)\times\{0\}$ which is not open in $\mathbb{R}^n\times\mathbb{R}^n$, right?

A:

You are correct that the image of a section $M \rightarrow TM$ is not open. One reason: since the section is a (smooth) immersion, the rank of the differential is no greater than the dimension of the domain, and here we have $\dim M < \dim TM$. Since the rank of the differential is less than that of the codomain, the image of the section must have dimension less than the codomain, and hence the image cannot be open (an open set must have dimension everywhere equal to the surrounding space).

