This question has also been posted on MSE, but maybe here is the right place to obtain an answer.


Let $(M^3,g)$ be a compact connected oriented Riemannian $3$-manifold with nonempty boundary. The Hodge-de Rham Theorem says that there is an isomorphism between $H^1\_{dR}(M)$, the first de Rham cohomology group of $M$, and the space $\mathcal{H}\_N^1(M)$ of tangential harmonic $1$-forms on $M$, given by


$$\mathcal{H}\_N^1(M) = \{ \omega \in \Omega^1(M) : d \omega = 0,\, d^\ast \omega=0 \text{ and } \omega(\nu) = 0 \text{ on }\partial M\},$$


where $\nu$ is a unit normal for $\partial M$.


Suppose that $H^1(M; \mathbb{Z}) \neq 0$. Then, since $\mathbb{S}^1$ is a $K(\mathbb{Z},1)$, there is a bijection


$$\Phi : [M, \mathbb{S}^1] \to H^1(M; \mathbb{Z})$$


given by $\Phi([u]) = [u^\ast(d\theta)]$, where $d\theta \in \Omega^1(\mathbb{S}^1)$ is the volume element of $\mathbb{S}^1$. Now, de Rham Theorem and the Universal Coefficient Theorem give isomorphisms


$$H\_{dR}^1(M) \cong H^1(M; \mathbb{R}) \cong H^1(M;\mathbb{Z}) \otimes\_{\mathbb{Z}} \mathbb{R}$$


How do I conclude from this information that there exists a smooth map $u : M \to \mathbb{S}^1$ such that $u$ is harmonic with Neumann condition and $u^\ast(d\theta)\in \Omega^1(M)$ is tangential and harmonic?



Tangential harmonic $1$-forms are pullbacks of harmonic functions
Any constant map satisfies the requirements of the final question :). More seriously, if you want to find tangential harmonic form of this type, which represents a given $[u]\in [M,S^1]\cong H^1(M,\mathbb{Z})$ in de Rham cohomology, then you can proceed as follows (unless I am missing something):


Pick a smooth map $v:M\rightarrow S^1$ representing $[u]\in [M,S^1]$. Then $\sigma:=v^\*d\theta$ is in the right de Rham cohomology class, if $\int\_{S^1}d\theta=1$. By the relative Hodge-de Rham theorem, there is an exact form $df\in \Omega^1(M)$ such that
$\tau:=\sigma+df$ is tangential and harmonic. Since we can write $df$ as $$df=(p\circ f)^\*d\theta$$ (where $p:\mathbb{R}\rightarrow S^1$ is a covering map with $p^\*d\theta=dt$, with $t$ as the standard coordinate on $\mathbb{R}$), we find that $\tau=(v+p\circ f)^\*d\theta$ has the desired form. Moreover $v+p\circ f$ is harmonic since $\tau$ is $d^\*$-closed and it satisfies the von Neumann condition since $\tau$ is tangential.


**Edit:** As requested in the comments, here are some more details on how one concludes that $v+p\circ f$ is harmonic:  

A map $w:M\rightarrow S^1$ is harmonic if and only if for all local isometries $L:S^1\supset U\rightarrow \mathbb{R}$ the function $L\circ w:w^{-1}(U)\rightarrow \mathbb{R}$ is harmonic. A real valued function $g$ is harmonic if and only if $dg=g^\*dt $ is $d^\*$-closed (as the Laplacian reduces on functions to $d^\*d$). If $g=L\circ w$, then $(L\circ w)^\*dt= c w^\*d\theta$ for a constant $c$. Combining these equivalences with the fact that $w^\*d\theta $ is $d^\*$-closed for $w=v+p\circ f$ we find that $v+p\circ f:M\rightarrow S^1$ is harmonic as desired.


