Continuity of a map to show that differential is full rank in some open set. I was reading a bit about differential geometry in Lee's book introduction to smooth manifolds. I came across the following proposition
Proposition 4.1. Suppose $F : M \to N$ is a smooth map and $p \in M$ . If $dF_p$ is
surjective, then $p$ has a neighborhood $U$ such that $F\mid_U$ is a submersion. If $dF_p$ is
injective, then $p$ has a neighborhood $U$ such that $F\mid_U$ is an immersion.
Proof. If we choose any smooth coordinates for $M$ near $p$ and for $N$ near $F(p)$,
either hypothesis means that Jacobian matrix of $F$ in coordinates has full rank at $p$.
Example 1.28 shows that the set of $m\times n$ matrices of full rank is an open subset of
$M(m\times n, \mathbb{R})$ (where $m = dim M$ and $n=dim N$ ), so by continuity, the Jacobian of
F has full rank in some neighborhood of p.
Here we are using the continuity of what map exactly? The map $p \mapsto dF_p$ right? The domain of the map is $M$ but what exactly is the codomain of this map? $dF_p \in \operatorname{Linear}(T_pM,T_pN)$ so the codomain would be $\bigcup_{p\in M} \operatorname{Linear}(T_pM,T_pN)$. It seems kind of complicated to me to show that this map is continuous.
 A: To avoid working with the map $p \mapsto dF_p$ whose codomain is space of linear maps, let's work with charts. Choose smooth charts $(V,\varphi, x^i)$ contain $p$ and $(W,\psi)$ contain $F(p)$. Denote $\hat{p} = \varphi(p)$. We actually considering the map from $\hat{V}=\varphi(V) \subseteq \mathbb{R}^m$ to space of matrices $M(m\times n,\mathbb{R})$,
$$
J: \hat{V} \rightarrow M(n\times m,\mathbb{R})
$$
defined by
$$
J : \hat{q} \longmapsto \Bigg[\frac{\partial \hat{F}^i}{\partial x^j}(\hat{q})\Bigg] \in M(n\times m,\mathbb{R}), \quad \forall \hat{q} \in \hat{V}.
$$
This map is smooth (hence continous) since each entries is a smooth functions of $\hat{q} \in \hat{V}$. By hypothesis $J(\hat{p})$ contain in an open subset of full rank matrices. Now you can argue by continuity of $J$.
A: The fact that $F:M^m \to N^n$ is $C^1$ means that you get a map $\varphi:U \to M_{n \times m }(\mathbb{R})$ given by $p \mapsto dF_p$, where $U$ is a small enough open neighborhood of $M$ that is contained in a smooth chart for $F$. $F$ is smooth so the map is continuous. If you take $p$ where $dF_p$ is injective\surjective you know there is a neighborhood  $ S \subset M_{n \times m }(\mathbb{R})$ of $dF_p$, of matrices of full rank. 
