Suppose $g : U \subseteq \mathbb{R}^m \to M \subseteq \mathbb{R}^k$, and $u \in U$ why does the derivative $dg_u$ extend to all of $\mathbb{R}^m$?
This is my question, it comes after reading page 4 of Topology from the Differentiable Viewpoint by Milnor
This is the excerpt :
"Now let us define the tangent space for an arbitrary smooth manifold $M \subseteq \mathbb{R}^k$. Choose a parameterization $$g : U \to M \subseteq \mathbb{R}^k$$ of a neighbourhood $g[U]$ of $x \in M$, with $g(u) = x$. Here $U$ is an open subset of $\mathbb{R}^m$. Think of $g$ as a mapping from $U$ to $\mathbb{R}^k$ so that the derivative $$dg_u : \mathbb{R}^m \to \mathbb{R}^k$$ is defined. Then $$TM_x = dg_u[\mathbb{R}^m]$$"
So my question is if $g$ maps $U$ to $m$, $g : U \to M$ how can its derivative at the point $u$, $dg_u$ extend to all of $\mathbb{R}^m$ so that it maps all of $\mathbb{R}^m$ into $\mathbb{R}^k$?
In other words if $g : U \to M$ how can $dg_u$ be defined so that $dg_u : \mathbb{R}^m \to \mathbb{R}^k$?
It seems that you're confusing the domain of $dg$ with that of $dg_u$. Remember that $dg_u$ is the derivative of $g : U \to M$ at the single point $u \in U$; i.e. the best linear approximation to $g(\cdot) - g(u)$ near $u$. You might be more familiar with this as the Jacobian matrix of the mapping $g$. Linear maps/matrices always map between vector spaces - in this case since $U \subset \mathbb R^m, M \subset \mathbb R^k$, the derivative is a $k \times m$ matrix, or equivalently a linear map $\mathbb R^m \to \mathbb R^k$.
If you want to talk about the derivative $dg$ as a function of position $u$, then you really have a map $dg : U \times \mathbb R^m \to \mathbb R^k$ defined by $dg(u,v) = dg_u(v)$.