I'm reading Warner. "Foundations of Differentiable Manifolds and Lie Groups." In the proof of corollary (a) in page 24, which I will present below, the book proves that the differential $df$ of the function $f: U \to \mathbb R^d$, where $U$ is open in $\mathbb R^d$, is injective. Then the book uses that fact to derive that the derivative $Df(m)$ of $f$ at $m$ is injective (or equivalently the Jacobian of $f$ at $m$ is non-singular). But I don't know why the injectivity of $df$ implies the injectivity of $Df(m)$.
I'll write here the corollary I'm referring to. But first I'll write the definitios of tangent vector and differential because the definitions might be different in other books.
Definition of tangent vector $\;$ A tangent vector $v$ at the point $m\in M$ is a linear derivation of the algebra $\tilde F_m$. That is, for all f, g$\in\tilde F_m$ and $\lambda\in\mathbb R$,
(a) $v($f$+\lambda$g$) = v($f$) + \lambda v($g$)$.
(b) $v($f$ \cdot $g$) =$ f$(m)v($g$) +$ g$(m)v($f$)$.$M_m$ denotes the set of tangent vectors to $M$ at $m$ and is called the tangent space to $M$ at $m$.
In the above definition, $\tilde F_m$ is the set of germs at $m$.
Definition of differential $\;$ Let $\psi: M\to N$ be $C^\infty$, and let $m\in M$. The differential of $\psi$ at $m$ is the linear map $$d\psi: M_m \to N_{\psi(m)}$$ defined as follows. If $v\in M_m$, then $d\psi(v)$ is to be a tangent vector at $\psi(m)$, so we describe how it operates on functions. Let $g$ be a $C^\infty$ function on a neighborhood of $\psi(m)$. Define $d\psi(v)(g)$ by setting $$d\psi(v)(g) = v(g\circ\psi).$$
In the above definition the author used the convention of writing $w($f$)$, where $w$ is a tangent vector and f is a germ of $f$, as $w(f)$.
Now I'll write the corollary and the proof I mentioned. The boldfaced part in the proof in what I don't understand
Corollary (a) $\;$ Assume that $\psi: M\to N$ is $C^\infty$, that $m\in M$, and that $d\psi: M_m\to N_{\psi(m)}$ is an isomorphism. Then there is a neighborhood $U$ of $m$ such that $\psi:U\to \psi(U)$ is a diffeomorphism onto the open set $\psi(U)$ in $N$.
Proof $\;$ Observe that $\operatorname{dim} M = \operatorname{dim} N$, say $d$. Choose coordinate systems (I think they are also called "charts" in other books) $(V, \varphi)$ about $m$ and $(W, \tau)$ about $\psi(m)$ with $\psi(V)\subset W$. Let $\varphi(m) = p$ and $\tau(\psi(m)) = q$. The differential of the map $\tau\circ\psi\circ\varphi^{-1} | \varphi(V)$ is non-singular at $p$. Thus the inverse function theorem yields a diffeomorphism $\alpha: \tilde U\to\alpha(\tilde U)$ on a neighborhood $\tilde U$ of $p$ with $\tilde U\subset\varphi(V)$.
Then $\tau^{-1}\circ\alpha\circ\varphi$ is the required diffeomorphism on the neighborhood $U = \varphi^{-1}(\tilde U)$ of $m$.
To use the inverse function theorem, one needs the fact that the Jacobian in non-singular, not the differential. But why does the author just proves that the differential is non-singular?
Note that $\tau \circ \psi \circ \varphi^{-1}$ is the map $\psi$ expressed in terms of local coordinates and is a smooth function from $\mathbb{R}^d$ to $\mathbb{R}^d$. Therefore, if you take the differential, which is a linear map, and you express it as a matrix with the bases induced by your choice of charts, the matrix will exactly be the Jacobian.