An integral map from 3-torus $\mathbb{T}^3$ to 3-sphere $S^3$ Let $\phi_1, \phi_2, \phi_3, \phi_4 \in \mathbb{R}$ be real valued functions, such that
$$\phi_j(x,y,z):(x,y,z) \in \mathbb{T}^3 \to \phi_j(x,y,z) \in  \mathbb{R}.$$
Here $\mathbb{T}^3$ is a 3-torus, with $j=1,2,3,4$.
The $\phi_j(x,y,z)$ satisfies a constraint
$$\sum_{j=1}^4 (\phi_j)^2=1,$$
which means that $(\phi_1, \phi_2, \phi_3, \phi_4)$ is a vector on a 3-sphere $S^3$.
Consider the integral computed from the domain $(x,y,z) \in\mathbb{T}^3$ to the target of $(\phi_1, \phi_2, \phi_3, \phi_4) \in S^3$. We can choose the  $\mathbb{T}^3$ has a unit length 1, and the $S^3$ has a unit radius 1.

Question 1:
Can we show that
$$(2/\pi^2) \int_{T^3} (\epsilon^{abc} \phi_1 \partial_a \phi_2 \partial_b \phi_3 \partial_c \phi_4) \;dx dy dz\;\in \mathbb{Z}?$$
is integer valued? (Or up to a front factor to be fixed.)
Is this true or is it wrong? (At least for certain function $\phi_j(x,y,z)$, I find the integral can be integer valued.



(Bonus, but you can skip this one below to claim the answer.)
Question 2:
More generally, is there some homotopy type of constraint, such that the integral map from the domain $\mathbb{T}^d$ to the sphere $S^d$, certain integral of the similar form
$$\# \int_{T^d} (\epsilon^{\mu_1 \mu_2 \mu_3 \dots \mu_d} \phi_1 \partial_1 \phi_2 \dots \partial_{\mu_{d-1}} \phi_{d} \partial_{\mu_d} \phi_{d+1}) \;d^dx \;\in \mathbb{Z}?$$
where $$\sum_{j=1}^d (\phi_j)^2=1,$$
Up to a proper normalization $\#$?

 A: Consider the three forms $\psi = x_1 dx_2\wedge dx_3 \wedge dx_4$. Write $\phi : \mathbb T^3 \to \mathbb R^4$, $\phi = (\phi_1, \cdots, \phi_4)$. Then 
\begin{align} 
\int_{\mathbb T^3} \phi^* \psi &= \int_{\mathbb T^3} \phi_1 d\phi_2 \wedge d\phi_3 \wedge d\phi_4 \\
&= \int_{\mathbb T^3} \epsilon^{abc} \phi_1 \partial_b \phi_2 \partial _c \phi_3 \partial _c\phi_4 \ \mathrm d x\  \mathrm d y\  \mathrm d z. 
\end{align}
On the other hand, 
$$\int_{\mathbb T^3} \phi^* \psi = \operatorname{deg} (\phi) \int_{\mathbb S^3} \psi, $$
where $\operatorname{deg}$ is the deg of the map $\phi$, which is an integer. Finally, by Stokes theorem, 
$$\int_{\mathbb S^3} \psi = \int_B d\psi = \int_B dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4.$$
The last term is the volume of the unit ball in $\mathbb R^4$ and is $\pi^2/2$. Thus your term equals $\operatorname{deg}(\phi)$. 
The generalization to the higher dimensional case should be easy. 
Edit To clarify, in general for two compact orientable $n$-dimensional manifold $M, N$, the degree of a smooth map $\phi : M\to N$ defined as 
$$ \int_M \phi^* \alpha = \operatorname{deg}(\phi) \int_N \alpha, \ \ \ \forall \alpha $$
is always an integer. I am following section 4 in Bott and Tu here. The above equality depends only on the cohomology class $[\alpha]$ instead of $\alpha$ itself. Thus we can assume $\alpha$ is a bump form support in a small open set around any point $q\in N$. Given a smooth $\phi$, let $q\in N$ be a regular value for $\phi$ (which exists by Sard's theorem). Then $\phi^{-1}(q)$ is a compact smooth submanifold of dimension $0$: that is, a finite set of points. Also there are open neighborhood of $q\in N$ so that $\phi : \phi^{-1}(B) \to B$ is a covering. Thus 
$$ \int_M \phi^* \alpha = \int_{\phi^{-1}(B)} \phi^* \alpha = \sum (\pm 1) \int_B\alpha $$
This $\sum (\pm 1)$ is the degree of $\phi$, you have $\pm 1$ since $\phi$ is a local diffeomorphism. 
