I am trying to prove that if $M$ is a $k$-manifold in $\mathbb R^n$, then $TM=\{(p, v): p \in M, v \in T_pM\}$ is a manifold. Here, $T_pM$ is defined as a subset of $\mathbb R^n$.
I know that questions about this problem have already been asked in this website, but my question is very specific:
Let $(p, v) \in TM$ be given. I attempted to create a coordinate patch as follows:
Suppose $\alpha: U\rightarrow M$ is a coordinate patch for $p$. Let $f: U \times \mathbb R^k\rightarrow TM$ be given by: $$f(x, u)=(\alpha (x), d\alpha(x)\cdot u)$$
I proved $f$ is $C^\infty$, injective, that $f[U \times \mathbb R^k]$ is a open set of $TM$. It remains to show that $f^{-1}$ is continuous.
Notice that $\alpha^{-1}(p, v)=(\alpha^{-1}(p), [d\alpha(\alpha^{-1}(p))]^{-1}\cdot v)$, where $[d\alpha(\alpha^{-1}(p))]^{-1}$ denotes the inverse of the linear map $d\alpha(\alpha^{-1}(p))$ as a function (we know that this function is injective, we set it's range, $T_pM$, as the inverse's domain).
I want to see that the function $(p, v)\rightarrow [d\alpha(\alpha^{-1}(p))]^{-1}\cdot v$ defined on $TM$ is continuous. I don't know how to proceed without some handwaving. Can someone clarify this?
Let $ \ F = G \circ d \alpha \circ \alpha^{-1} \ $, where $ \ G(T)=T^{-1}$. Then $F$ is continuous. If $ \ (p_0,v_0) \in TM \ $ and $ \ \varepsilon > 0 \ $ there exists $ \ \delta > 0 \ $ such that, $\forall p \in U$, $$\Vert p-p_0 \Vert < \delta \ \Rightarrow \Vert F(p) - F(p_0) \Vert < \sqrt{\varepsilon} \ . $$
Let $ \ \eta = min \{ \delta , \sqrt{\varepsilon} \}$. If $ \ \ max \{ \Vert p-p_0 \Vert , \Vert v-v_0 \Vert \} = \Vert (p,v) - (p_0,v_0) \Vert < \eta$, we have $$\Vert [F(p)](v) - [F(p_0)](v_0) \Vert = \Vert [F(p)-F(p_0)](v-v_0) \Vert \leqslant \Vert F(p-p_0)\Vert \cdot \Vert v-v_0 \Vert <$$ $$< \sqrt{\varepsilon} \cdot \Vert v-v_0 \Vert < \sqrt{\varepsilon} \cdot \sqrt{\varepsilon} = \varepsilon \ . $$