Let $(M^m,g)$ and $(N^n,h)$ be Riemannian manifolds. A smooth map $\pi : M \to N$ is called a harmonic morphism if for any harmonic function $f : U \to \mathbb{R}$ defined on an open set $U \subseteq N$ such that $\pi^{-1}(U) \neq \emptyset$, it holds that $f \circ \pi : \pi^{-1}(U) \to \mathbb{R}$ is a harmonic map.
The following theorem is useful for verifying whether a map is a harmonic morphism:
Theorem. A non-constant map $\pi : M^m \to N^n$, $m \geq n$, is a harmonic morphism if and only if it is a harmonic map and horizontally conformal.
By horizontally conformal I mean the following. For $x \in M$, let $V_x = \ker d \pi(x) \subseteq T_x M$ and $H_x = V_x^\perp$. Also let $C_\pi = \{ x \in M : d\pi(x) = 0\}$ and $\hat{M} = M \setminus C_{\pi}$. Then $\pi$ is horizontally conformal when there exists a function $\lambda : \hat{M} \to \mathbb{R}_+$ such that $$\lambda(x)^2 g(X,Y) = h(d\pi(X), d\pi(Y))$$ for every $X,Y \in T_x M$ and $x \in \hat{M}$.
It is easy to see that if $\pi : \tilde{M} \to M$ is a Riemannian covering map, then it is horizontally conformal. Moreover, Eells and Sampson showed that every covering map is harmonic. So, we conclude that $\pi : \tilde{M} \to M$ is a harmonic morphism.
Now comes my question: Let $\pi : \tilde{M} \to M$ as above and let $f : M \to \mathbb{S}^1 = \mathbb{R}/\mathbb{Z}$ be a harmonic map. Is it true that $f \circ \pi : \tilde{M} \to \mathbb{S}^1$ is a harmonic map?