Let $S = \{f:X\times \mathbb R^n \to \mathbb R^n : f\ \text{continuous}, \ f(x,\_) \ \text{linear map}\}$, $T = \{g:X \to M_n(\mathbb R) : g\ \text{continuous}\}$ with $X$ a topological space, prove that exists a bijection between $S$ and $T$
Ok, I know that exists a bijection between
$\tilde{S} = \{f:X\times \mathbb R^n \to \mathbb R^n : f\ \text{function}, \ f(x,\_) \ \text{linear map}\}$ $\ $ and $\ $ $\tilde{T} = \{g:X \to M_n(\mathbb R) : g\ \text{function}\}$ given by $\varphi(f) (x)(a) = f(a,b)$ with inverse $\psi(g) (x,a) = g(x)(a)$.
(I really use this bijection between $S$ and $\overline{T}$ with $\overline{T} = \{g:X \to Hom_{\mathbb R}(\mathbb R^n) : g\ \text{continuous}\}$ , but $Hom_{\mathbb R}(\mathbb R^n) \simeq M_n(\mathbb R)$ so there is no problem here.)
My problem is proving that if $f$ is continuous, then $\varphi(f)$ is continuous and if $g$ is continuous, then $\psi(g)$ is continuous
Since the bijection $C(X, C(\mathbb{R^n}, \mathbb{R^n}))\cong C(X\times \mathbb{R^n}, \mathbb{R^n})$ is true with $f:X\times\mathbb{R}^n\to\mathbb{R}^n$ mapping to $g(x)(y) = f(x, y)$ and $C(\mathbb{R}^n, \mathbb{R}^n)$ equipped with compact-open topology, its clear that it induces a bijection$$C(X, M_n(\mathbb{R}))\cong \{f\in C(X\times \mathbb{R}^n, \mathbb{R}^n) : \forall_{x\in X}\ y\mapsto f(x, y)\text{ is linear}\}$$ with $M_n(\mathbb{R})\subseteq C(\mathbb{R}^n, \mathbb{R}^n)$ having subspace topology. Thus its enough to show that compact-open topology on $M_n(\mathbb{R})$ is the same as the standard one:
Note that the inclusion $M_n(\mathbb{R})\to C(\mathbb{R}^n, \mathbb{R}^n)$ with $M_n(\mathbb{R})$ having standard topology is continuous since the map $M_n(\mathbb{R})\times \mathbb{R}^n\to \mathbb{R}^n$ given by $(A, x)\mapsto Ax$, multiplication of vector by a matrix, is continuous. Moreover, a basic open set in $M_n(\mathbb{R})$ is of the form $$U = \{B\in M_n(\mathbb{R}) : \forall_{i=1, ..., n}\|Ae_i-Be_i\| < \varepsilon\}$$ where $A\in M_n(\mathbb{R})$ and $e_1, ..., e_n$ is the canonical basis on $\mathbb{R}^n$. Then $U = M_n(\mathbb{R})\cap \bigcap_{i=1}^n V_i$ where $$V_i = \{f\in C(\mathbb{R}^n, \mathbb{R}^n) : \|f(e_i)-Ae_i\| < \varepsilon\}$$ is an element of the subbasis of $C(\mathbb{R}^n, \mathbb{R}^n)$ with compact set $\{e_i\}$ and open set $\{y : \|y-A(e_i)\|< \varepsilon\}$. Thus the inclusion $M_n(\mathbb{R})\to C(\mathbb{R}^n, \mathbb{R}^n)$ is actually an embedding so that the two topologies on $M_n(\mathbb{R})$ coincide.