The short answer is yes.

Strictly speaking, the definition of a factor of a measure preserving system (m.p.s.) is as follows.

**Definition:** Let $(X,\mathcal{B},\mu,T)$ and $(Y,\mathcal{C},\nu,S)$ be m.p.s. Then, a set
function $\phi: X'\to Y'$ is called a factor map if it is surjective, measure preserving
in the sense that $\mu(\phi^{-1}A)=\nu(A)$, for any $A \in \mathcal{C}$, and satisfies the equality
$$S\circ \phi (x) = \phi \circ T(x),\ \text{for}\ x \in X',$$
where $X'$ and $Y'$ are full measure subsets of $X$
and $Y$ respectively, with $TX' \subseteq X'$ and $SY' \subseteq Y'$. We say that the system
$(Y,\mathcal{C},\nu,S)$ is a factor of $(X,\mathcal{B},\mu,T)$, if such a factor map exists and in this case we also say
that $(X,\mathcal{B},\mu,T)$ is an extension of $(Y,\mathcal{C},\nu,S)$.

**''Equivalent'' definitions:** There are two more equivalent ways of seeing factors of m.p.s. one through $T$-invariant sub-$\sigma$-algebras and one through $T$-invariant
algebras of bounded measurable functions.

Let $(X,\mathcal{B},\mu,T)$ be a m.p.s. and $(Y,\mathcal{C},\nu,S)$ some factor of it, with factor
map $\phi: X'\to Y'$. Then, it is easy to see
that $\phi^{-1}(\mathcal{C})$ is a $T$-invariant sub-$\sigma$-algebra of $\mathcal{B}$ (modulo sets of measure zero).
For example, if $A \in \phi^{-1}(\mathcal{C})$, then $A=\phi^{-1}(C)$, for some $C\in \mathcal{C}$ and so
$$T^{-1}A=(\phi \circ T)^{-1}(C)=(S \circ \phi)^{-1}(C)=\phi^{-1}(S^{-1}(C)) \in \phi^{-1}(\mathcal{C}).$$

A partial converse is also true. Precisely, if $(X,\mathcal{B},\mu,T)$ is some
m.p.s. and $\mathcal{A} \subset \mathcal{B}$ is a $T$-invariant sub-$\sigma$-algebra then,
there exists a system $(Y,\mathcal{C},\nu,S)$ and a factor map $\phi: X \to Y$, such that
$\mathcal{A}= \phi^{-1}(\mathcal{C})$.

We can also describe a correspondence between invariant $\sigma$-algebras and
invariant algebras of functions. Indeed, if $(X,\mathcal{B},\mu,T)$ is a m.p.s. and $\mathcal{F}$ is
a $T$-invariant algebra of $L^{\infty}(\mu)$, that is, $T\mathcal{F} \subset \mathcal{F}$ and for any $f,g \in \mathcal{F}$
and $c\in \mathbb{C}$, we have that $f\cdot g, f+cg \in \mathcal{F}$, the set
$$\mathcal{A}:=\{B\in \mathcal{B}: B=f^{-1}(D),\ \text{for some}\ f\in \mathcal{F}\ \text{and}\ D\ \text{a complex Borel set}\}$$
is a $T$-invariant sub-$\sigma$-algebra of $\mathcal{B}$. On the other hand, given $\mathcal{A} \subset \mathcal{B}$ some
$T$-invariant sub-$\sigma$-algebra of $\mathcal{B}$, the set
$$\mathcal{F}:=\{f\in L^{\infty}(\mu): f\ \text{is measurable with respect to}\ \mathcal{A}\}$$
is a $T$-invariant sub-algebra of $L^{\infty}(\mu)$.

**Conclusion:** So, to conclude answering your question, we can define the Kronecker factor through functions; i.e. as the closed linear subspace of $L^2(\mu)$
$$\mathcal{K}_T:=\overline{\text{span}\{f\in L^{\infty}(\mu):f\ \text{is an eigenfunction of}\ T\}}^{L^2(\mu)},$$
where, a function $f\in L^{\infty}(\mu)$ is an
eigenfunction of $T$ if there exists $a\in \mathbb{R}$, so that $Tf=e^{2\pi ia}f$ and $e^{2\pi ia}$ is the
respective eigenvalue, we can define it as the minimal $T$-invariant sub-$\sigma$-algebra of $\mathcal{B}$ with respect to which all eigenfunctions are measurable, or finally, we can define it as the maximal factor of the system $(X,\mathcal{B},\mu,T)$ that is isomorphic to a rotation on a compact abelian group.