The quotients GL_n/O(n) and SL_n/SO(n)

The map. Define $\pi:GL_n\to\mathrm{PD}$, $\pi(A)=AA^T$. It is continuous, and for $U\in O(n)$,
$$\pi(AU)=AUU^TA^T=AA^T=\pi(A),$$
so $\pi$ is constant on left cosets $A\cdot O(n)$ and induces a continuous map $\bar\pi:GL_n/O(n)\to\mathrm{PD}$. (Note: with $A\mapsto A^TA$ one would get the space of right cosets $O(n)\backslash GL_n$.)

Bijective. *Surjective:* for $P\in\mathrm{PD}$, $A=P^{1/2}$ satisfies $AA^T=P$. *Injective:* if $AA^T=BB^T$, put $U=B^{-1}A$; then $UU^T=B^{-1}AA^TB^{-T}=B^{-1}BB^TB^{-T}=I$, so $U\in O(n)$ and $A=BU\in B\cdot O(n)$.

Continuous inverse. The inverse is $P\mapsto[P^{1/2}]$, the composition of the continuous map $P\mapsto P^{1/2}=\exp(\frac12\log P)$ and the quotient map $GL_n\to GL_n/O(n)$. So $\bar\pi$ is a homeomorphism.

The case of $SL_n$. Restricting, $SL_n/SO(n)\to\{P\in\mathrm{PD}:\det P=1\}$ is a continuous bijection: if $A\in SL_n$ then $\det AA^T=1$; conversely, for $\det P=1$ the square root $P^{1/2}$ has determinant $1$ and lies in $SL_n$. Also $AA^T=BB^T$ with $A,B\in SL_n$ gives $U=B^{-1}A\in O(n)$ with $\det U=1$, so $U\in SO(n)$.

Identification with a vector space. By $\det e^S=e^{\operatorname{tr}S}$ and the bijection $\exp:\mathrm{Sym}_n\to\mathrm{PD}$, $\det e^S=1$ iff $\operatorname{tr}S=0$. So $\exp$ restricts to a homeomorphism
$$\{S\in\mathrm{Sym}_n:\operatorname{tr}S=0\}\ \to\ \{P\in\mathrm{PD}:\det P=1\},$$
and the space of traceless symmetric matrices is a vector space of dimension $\frac{n(n+1)}2-1$.

Check. $n=2$: $SL_2/SO(2)\cong\mathbb{R}^2$, the hyperbolic plane (as a topological space an open disc).

Remark. Another consequence: $GL_n$ deformation retracts onto $O(n)$ and $SL_n$ onto $SO(n)$, so topologically $SL_n\simeq SO(n)$.