$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be closed $d$-dimensional oriented Riemannian manifolds. Let $f:\M \to \N$ be smooth, and let $\delta=d^*$ be the adjoint of the exterior derivative.
Let $1 \le k \le d$ be fixed. Consider the following two properties $f$ can have:
- $\delta^{\N} \omega=0 \Rightarrow \delta^{\M}(f^*\omega)=0$ for every $k$-form $ \omega \in \Omega^k(\N)$.
$\,\,\,\,$ 2. $\omega\in \Omega^k(\N)$ is harmonic $\Rightarrow f^*\omega$ is harmonic.
Question: Does property 2 implies property 1?
(Property 1 certainly implies property 2, since a form is harmonic if and only if it's closed and co-closed, and closedness of forms is preserevd automatically, by any smooth map).
Comment: A map which satisfies property $2$ for $k=1$, is called harmonic morphism.
I suspect the answer is negative, since the space of harmonic forms is finite-dimensional. Thus property $2$ gives us a "finite-dimensional" information about $f$, while the requirement of property $1$ is on the much larger space $\text{ker} \delta$, which is infinite dimensional.
Since you've phrased the question in terms of global forms, the answer is definitely no. For example, let $M=N=\mathbb S^2$. Then by the Hodge theorem, there are no nontrivial harmonic $1$-forms, so every smooth map $f\colon \mathbb S^2\to\mathbb S^2$ satisfies property 2 in the case $k=1$. But not every such $f$ satisfies property 1. (I was thinking that $f\colon \mathbb S^2\to\mathbb S^2$ satisfies property 1 for $k=1$ if and only if it's holomorphic or antiholomorphic, but I don't believe that's true.)