I'm sure you'll get rigorous answers on this site as well, but you wouldn't be here if the highly rigorous definition you did not appreciate hadn't defined itself out of unambiguous meaning. I'll work by example which might or might not satisfy you, but that's the point: define something so it makes some sense to most bona-fide seat-of-the-pants users.
Start with the most trivial case of one degree of freedom, in this case x, the real line. The wavefunction $\psi(x)$ is a complex normalized ($\int\!\!dx ~ \psi^*(x) \psi(x)=1$) function which serves to produce a probability measure, among other things, but no matter. x is the location on that line, the "observable", and there is no other observable operator commuting with $\hat x$, the operator associated with it. (E.g. the momentum operator $\hat p$ does not commute with it.) The expectation of $\hat x$ for that state is $\int\!\!dx ~ \psi^*(x) \psi(x) ~x=\langle \hat x\rangle$. You may think of $\psi(x)$ as a complex infinite-dimensional vector. In the spectacularly more meaningful language of rigged Hilbert space, it is but the expansion coefficients in the unnormalizable $|x\rangle$ basis, of the abstract state $|\psi\rangle$,
$$
|\psi\rangle= \int\!\!dx~ \psi(x) |x\rangle, ~~\leadsto\\
\psi(x)=\langle x|\psi\rangle, ~~\leadsto\\
\langle x|\hat x|\psi\rangle= x\psi(x).
$$
You may proceed to 3-dimensional space, $\mathbb R ^3$, where the observables $\hat x, \hat y, \hat z$ further commute among themselves. The variables are now 3-vectors, x, the integrals are 3-dimensional, etc., 3 degrees of freedom. Hardly any topology involved.
Now here comes a twist, all but guaranteed to drive mathematicians mad. You may choose to focus on further observables, e.g., $\psi ({\mathbf x}; E, \ell, m)$, commuting among themselves, but not with x, separated by the semicolon in the wavefunction. They are the eigenvalues E of $\hat H$, $\hbar^2 \ell (\ell +1)$ of $ L^2$, and ℏm of $L_z$ in atoms. That is,
$$
\langle {\mathbf x}|\hat H |\psi\rangle = E~\psi ({\mathbf x}; E, \ell, m),\\
\langle {\mathbf x}|\hat L_z/\hbar |\psi\rangle = m~\psi ({\mathbf x}; E, \ell, m),\\
\langle {\mathbf x}|\hat L^2/\hbar^2 |\psi\rangle = \ell (\ell+1)~\psi ({\mathbf x}; E, \ell, m).
$$
This is to say the wavefunction corresponds to the state vector $|\psi\rangle$ which is an eigenstate of the mutually commuting $\hat H$, $\hat L^2$, and $\hat L_z$, but, even as these do not commute with $\hat {\mathbf x}$, the wavefunction corresponding to that state is the unambiguous expression above; a bit of a centaur, based on the projection of $|\psi\rangle$ on $|x\rangle$. $|\psi\rangle$ is not an eigenstate of $\hat x$, but only of the three observables after the semicolon, but the wavefunction is! A visitor from another field is justified to be puzzled, and you might improve the wikipedia presentation when you find your footing. You might sort out any topological issues, if you believed there are any.