Help understanding a little Jacobians lemma This was used as lemma in a bigger proof.

Let $A ⊆ ℝ^m$ be an open set, $f = (f_1,…,f_m) : A → ℝ^m$ a $C^1$ function, and $p ∈ A$. If $det[Df(p)] ≠ 0$, then there is an open ball $B ⊆ A$ of center $p$ such that $f|_B$ is injective.
proof:
The function $x ↦ det[Df(x)]$ is continuous, so there is an open ball $B ⊆ A$, of center $p$, such that $x ∈ B ⇒ det[Df(x)] ≠ 0$. Let $a$ and $b$ be two points of $B$. By the mean value theorem we have $f_i(b) - f_i(a) = Df_i(c_i)⋅(b-a)$ for some $c_i ∈ \{ta + (1-t)b : t \in [0,1]\}$, this means that $$f(b) - f(a) = \begin{bmatrix}Df_1(c_1)\\\vdots\\Df_m(c_m)\end{bmatrix}⋅(b-a).$$But $det[Df(c_i)] ≠ 0$ so $f(b) = f(a) ⇒ b = a$.

The last step is where I have troubles, it seems to me that we need $$det\begin{bmatrix}Df_1(c_1)\\\vdots\\Df_m(c_m)\end{bmatrix} ≠ 0$$
and it's not clear how this follows from $det[Df(c_i)] ≠ 0$.
 A: You are right, the last step needs to be justified.
I'm denoting with $|⋅|$ the max norm, so $|[a_{ij}]| = max_{i,j}|a_{ij}|$, and I'll call $T$ the matrix $\begin{bmatrix}Df_1(c_1)\\\vdots\\Df_m(c_m)\end{bmatrix}$. 
Being $det[Df(p)] ≠ 0$, we can write:
$$|b-a| = |Df(p)^{-1}Df(p)(b-a)| ≤ m|Df(p)^{-1}||Df(p)(b-a)| = σ^{-1}|Df(p)(b-a)| ⇒ \\|Df(p)(b-a)| ≥ σ|b-a|.$$
The function $x ↦ Df(x)$ is continuous, so we can choose the radius of $U$ is small enough that: $$x ∈ U ⇒ |Df(p)-Df(x)| < σ/m ⇒ |Df_i(p)-Df_i(c_i)| < σ/m ⇒ |Df(p)-T| < σ/m.$$
Now, if $b ≠ a$, we have:
$$σ|b-a| -  |T(b-a)| ≤ |Df(p)(b-a)| - |T(b-a)| ≤ |[Df(p) - T](b-a)| ≤ \\m|Df(p) - T||b-a| < σ|b-a| ⇒ |f(b) - f(a)| = |T(b-a)| > 0.$$
A: The proof does not seem to make sense because of an issue it has. First, note
that for the last step, the equation should read
$$
  f(b) - f(a) =
  \begin{bmatrix}
    Df_{1}(c) \\
    Df_{2}(c) \\
    \dots \\
    Df_{m}(c)
  \end{bmatrix}
  (b - a)
$$
where $c$ is the vector with components $c_{i}$. That is, each $f_{i}$ is a function from $R^{m}$ to $R^{1}$, so its linearization should have $R^{m}$ as the domain. Therefore, the previous claim 
 $$ f_{i}(b) - f_{i}(a) = Df_{i}(c_{i})(b - a) $$
needs to be corrected somehow. That should be sufficient to fix the proof.
