Let $M \subset \mathbb{R}^k$ be a smooth $m$-dimensional manifold and fix a point $p$ on $M$.
Let $U_0 \subset M$ be an $M$-open set with $p \in U_0$ and let $\phi_0 : U_0 → \Omega_0$ be a diffeomorphism onto an open subset $Ω_0 ⊂ \mathbb{R}^m$.
Define $x_0 := \phi_0(p)$ and let $\psi_0 := \phi_0^{-1}: \Omega_0 \rightarrow U_0$ be the inverse map. Then:
$$T_pM = (\text{im }d\psi_0(x_0): \mathbb{} \mathbb{R}^m \rightarrow \mathbb{R}^k) $$
An inclusion argument is trivial to make, i.e. for a small enough $t$ and $\xi \in \mathbb{R}^m$, such that $x_0 + t \xi \in \Omega_0$, we can define a curve $\gamma: (-\epsilon, \epsilon) \rightarrow M$, such that $\gamma(t) := \psi_0(x_0 + t\xi)$. Naturally, we get:
$$\gamma(0) = p, \text{ } \dot{\gamma}(0) = \frac{d}{dt} \bigg|_{t=0} \psi_0(x_0 + t\xi) = d\psi_0(x_0) \xi$$
This shows that $d\psi_0\xi \in T_pM$, which is all fine. The author then goes to say that since $\dim \text{im }d\psi_0(x_0) = m$, the problem is solved.
I'm missing the justification behind this $\dim \text{im }d\psi_0(x_0) = m$. Indeed, given the smoothness of $\psi_0 := \phi_0^{-1}$ I assume some argument can be made regarding the rank/surjectivity of the Jacobian $d\psi(x_0)$ via some form of the inverse function theorem, but I just can't get my finger on it.
Otherwise, regarding $d\psi(x_0)$ as the best linear approximation to $\psi$ around $x_0$, one may write:
$$\psi(x_0 + t \xi) = \psi(x_0) + td\psi(x_0)\xi + o(\Vert \xi \Vert)$$
But given $m \neq k$ I find it non-trivial to argue for the surjectivity of $d\psi(x_0)$. A solution would be welcome.
The text I'm studying is Introduction to Differential Geometry, Robin and Salamon (18 April 2020 version), page 26.
We follow J. Milnor, Topology from the differentiable viewpoint, page 5, almost verbatim.
Since $\phi_0\colon U_0\to\Omega_0$ is a smooth mapping, we can choose an open set $W$ containing $p$ and a smooth map $F\colon W\to\mathbf{R}^m$ that coincides with $\phi_0$ on $W\cap U_0$. Letting $V=\phi_0(W\cap U_0)$ and writing $\iota\colon V\to\mathbf{R}^m$ for the inclusion map, we have $\iota=F\circ\psi_0$ and therefore $\text{id}_{\mathbf{R}^m}=dF_p\circ d(\psi_0)_{x_0}$. It follows that $d(\psi_0)_{x_0}$ has rank $m$, and hence its image $T_pM$ has dimension $m$.