Trying to prove that $TM$ is a manifold: Is this function an homeomorphism? 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?

 A: 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 $$\Vert [F(p)](v) - [F(p_0)](v_0) \Vert = \Vert [F(p)-F(p_0)](v-v_0) \Vert \leqslant \Vert F(p-p_0)\Vert \cdot \Vert v-v_0 \Vert <$$ $$< \sqrt{\varepsilon} \cdot \Vert v-v_0 \Vert < \sqrt{\varepsilon} \cdot \sqrt{\varepsilon} = \varepsilon \ . $$
A: I assume you are talking about the tangent bundle as a subset of $M \times \mathbb{R}^n$, right?
Let me offer two alternative ways to solve this.

Firstly, since you are on a coordinate patch, you have a continuous (actually, smooth) way of choosing $n-k$ normal vector fields to your manifold which are pointwise linearly independent. Use them to extend your linear maps $d\alpha$ to invertible linear maps
with domain on $\mathbb{R}^n$ instead of $\mathbb{R}^k$. It is now easier to verify continuity (and smoothness) of the inverse map of this enlarged function $U \times \mathbb{R}^n \to M \times \mathbb{R}^n$, since we are dealing with bona fide invertible linear maps, not only injective ones. The function you want to be continuous is just the restriction of this to $TM$.

Secondly, assuming nothing besides topology, you can use this lemma:

Lemma: Let $X,Y$ be topological spaces, where $Y$ is Hausdorff, such that $X$ has an exhaustion by compact sets $K_n$, and let $F:X\rightarrow Y$ be a continuous bijective map such that $F(K_n)$'s exhaust $Y$. Then $F$ is a homeomorphism.
Proof: $F$ is by hypothesis continuous, and being injective on compact sets (the codomain is Hausdorff), each $F|_{K_n}$ is a homeomorphism with its image. We will prove that $G:=F^{-1}$ is continuous.
Take $y \in Y$, together with a open neighbourhood $W$ of $G(y)$. Since $K'_n:=F(K_n)$ exhaust $Y$, there exists a $K'_N$ such that $y \in K'_N$. By hypothesis, $y \in \text{int}K'_{N+1}$. Due to our previous observation, $G|_{K'_{N+1}}$ is a homeomorphism. Therefore, $G|_{\text{int}K'_{N+1}}$ is continuous (in fact, a homeomorphism). Hence, there exists an open neighbourhood $V$ of $y$ such that $G(V) \subset W$. But a neighbourhood in the induced topology of an open set must be an open set in the topology itself. Hence, we found an open neighbourhood of $y$ which is taken inside $W$. This proves continuity of $G$.

Considering that $U$ is open in $\mathbb{R}^k$, it has an exhaustion by compact sets $K_n$. The exhaustion $K_n \times [-n,n]^k$ of $U \times \mathbb{R}^k$ should satisfy the hypothesis of the lemma.
OBS: By exhaustion, I mean a sequence of compact sets such that $K_n \subset \text{int} K_{n+1}$.
