What does it mean to say that $Df : U → \text{Hom}(V, W)$ is continuous? The following is from this paper by K. Conrad.

What does it mean to say that $Df : U → \text{Hom}(V, W)$ is continuous? Upon fixing linear coordinates on $V$ and $W$, such continuity amounts to continuity for each of the component functions $∂x_jf_i: U → R$ of the matrix-valued $Df$, and so the concrete definition of $f$ being $C^1$ (namely, that each $∂x_jf_i$ exists and is continuous on $U$) is equivalent to the coordinate-free property that $f : U → W$ is differentiable and that the associated total derivative map $Df : U → \text{Hom}(V, W)$ from $U$ to a new vector space $\text{Hom}(V, W)$ is continuous.


Conrad jumps straight up to the statement that the continuity of $Df$ "amounts to continuity for each of the component functions $∂x_jf_i: U → R$", without providing an argument as to why this is the case. I have not worked a lot with $\text{Hom}(V,W)$ and so I struggle to understand the quoted paragraph. In particular:

*

*How can one define a topology on $\text{Hom}(V,W)$?


*Why is the continuity of $Df$ (treating $\text{Hom}(V,W)$ with the topology defined above) equivalent to the continuity of each $∂x_jf_i: U → R$?
 A: The topology of $\mathrm{E} = \mathsf{Hom}(\mathrm{V}, \mathrm{W})$ is given by the canonical norm
$$
\|u\| = \sup_{\|x\|_\mathrm{V} \leq 1} \|u(x)\|_\mathrm{W}.
$$
In this way, if $u, v \in \mathrm{E},$ then $\|u - v\|$ is the distance and thus a topology is born. As mentioned in the comments, $\mathrm{E}$ is finite dimensional when both $\mathrm{V}$ and $\mathrm{W}$ are finite dimensional, by virtue of the Bolzano-Weierstrass theorem, all norms on $\mathrm{E}$ induce the same topology for such theorem entails that if $\|u\|_1$ and $\|u\|_2$ denote two norms on $\mathrm{E}$ then there exists constants $a > 0$ and $b > 0$ such that for all $u \in \mathrm{E},$ $a \|u\|_1 \leq \|u\|_2 \leq b\|u\|_1.$ In this way, we can change the norm of $\mathrm{E}$ to suit our needs. The basic way this works is to choose bases of $\mathrm{V}$ and $\mathrm{W}$ and then induce coordinate (so we identify there spaces with $\mathbf{R}^p$ and $\mathbf{R}^q$ respectively) and this allows to induce a matrix structure on $\mathrm{E}$ by identifying it with $\mathsf{Mat}_{q \times p}$ which is also identified with $\mathbf{R}^{pq},$ so then $u = (u_{ij})$ is a matrix. It can then be shown that $u:\mathrm{U} \to \mathrm{E}$ (say $u:x \to (u_{ij}(x))$) is continuous if and only if its coordinate functions are, when $u = f',$ these coordinates are none other than $\partial_{x_i} f_j.$
