Compact open topology on $\operatorname{GL}(n, \mathbb{R})$ coincides with Euclidean topology. There are two ways to assign $\operatorname{GL}(n,\mathbb{R})$ topologies: as subspace of $\mathbb{R}^{n^2}$, or subspace of $\operatorname{Maps}(\mathbb{R}^n, \mathbb{R}^n)$  where the latter is given compact open topology.
I was reading Proposition 1.4, that these two coincides. I don't understand the proof except

On the one hand, the universal property of mapping space, Proposition 8.45, gives the inclusion is continuous,
$$
\operatorname{GL}(n, \mathbb{R}) \to \operatorname{Maps}(\mathbb{R}^n, \mathbb{R}^n)
$$

I don't understand—how? In fact I don't know what the universal property is.

The universal property of the compact-open topology ("mapping space") as in your link means nothing else than that for locally compact $Y$
(1) the evalation map $e : Z^Y \times Y \to Z, e(f,y) = f(x)$ is continuous
(2) the exponential correspondence $E : Z^{X \times Y} \to (Z^Y)^X$ is a bijection
To prove that $i : GL(n, \mathbb{R}) \to Maps(\mathbb{R}^n, \mathbb{R}^n)$ is continuous, it therefore suffices to show that $\alpha = E^{-1}(i) : GL(n, \mathbb{R}) \times \mathbb{R}^n \to \mathbb{R}^n$ is continuous. $\alpha$ is the restriction of the bilinear map $\tilde{\alpha} : End(\mathbb{R}^n) \times \mathbb{R}^n \to \mathbb{R}^n, \tilde{\alpha}(\phi,x) = \phi(x)$, where $End(\mathbb{R}^n)$ denotes the vector space of all endomorphisms of $\mathbb{R}^n$. But bilinear maps are continuous with respect to the Euclidean topologies (all occurring vector spaces are finite-dimensional).
