Vtyjkyuk

There are two ways to assign $operatornameGL(n,mathbbR)$ topologies: as subspace of $mathbbR^n^2$, or subspace of $operatornameMaps(mathbbR^n, mathbbR^n)$ 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,
$$
operatornameGL(n, mathbbR) to operatornameMaps(mathbbR^n, mathbbR^n)
$$

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 $i : GL(n, mathbbR) to Maps(mathbbR^n, mathbbR^n)$ is continuous, it therefore suffices to show that $alpha = E^-1(i) : GL(n, mathbbR) times mathbbR^n to mathbbR^n$ is continuous. $alpha$ is the restriction of the bilinear map $tildealpha : End(mathbbR^n) times mathbbR^n to mathbbR^n, tildealpha(phi,x) = phi(x)$, where $End(mathbbR^n)$ denotes the vector space of all endomorphisms of $mathbbR^n$. But bilinear maps are continuous with respect to the Euclidean topologies (all occurring vector spaces are finite-dimensional).