Poincare-Miranda on a torus The Poincare-Miranda theorem involves $d$ functions on a $d$-dimensional cube. Function $i$ is negative in one direction of dimension $i$ and positive in the opposite direction of dimension $i$. The theorem says that there must be a point in the cube in which all functions are 0.
Is there a parallel theorem on a $d$-dimensional torus, where the boundary conditions are replaced by oddity?
To be concrete, let $d=2$. A 2-dimensional torus is a cartesian product of 2 circles: $S^1 \times S^1$.
There are 2 continuous functions: $f_1(x_1,x_2), f_2(x_1,x_2)$, where $x_i$ is a point on the circle in dimension $i$. Function $i$ is odd in dimension $i$, i.e: $f_1(-x_1,x_2)=-f_1(x_1,x_2)$, and similarly $f_2(x_1,-x_2)=-f_2(x_1,x_2)$. 
(Alternatively, the two functions can be defined as: $f_1(\theta_1,\theta_2), f_2(\theta_1,\theta_2)$, where $\theta_i$ is an angle on the circle in dimension $i$. Function $i$ is odd in dimension $i$, i.e: $f_1(\theta_1+\pi,\theta_2)=-f_1(\theta_1,\theta_2)$ and $f_2(\theta_1,\theta_2+\pi)=-f_2(\theta_1,\theta_2)$).
Is this true that there must be a point in the torus in which all functions are 0?
The intuition for this variant is the same as for the original variant: for each function $i$, there must be a "wall" of 0 which separates the positive side of $i$ from its negative side. Each wall is in a different dimension, so they must meet in a point. But is it true?
 A: $\newcommand{\Reals}{\mathbf{R}}$View the unit circle as the set of unit complex numbers, and define a map $f:S^{1} \times S^{1} \to \Reals^{2}$ by
$$
f(e^{i\theta_{1}}, e^{i\theta_{2}})
  = \bigl(\underbrace{\cos(\theta_{1} + \theta_{2})}_{f_{1}(\theta_{1}, \theta_{2})},
          \underbrace{\sin(\theta_{1} + \theta_{2})}_{f_{2}(\theta_{1}, \theta_{2})}\bigr).
$$
For all $\theta_{1}$ and $\theta_{2}$,
\begin{align*}
f_{1}(\theta_{1} + \pi, \theta_{2})
  &= \cos(\theta_{1} + \theta_{2} + \pi)
  = -\cos(\theta_{1} + \theta_{2}), \\
f_{2}(\theta_{1}, \theta_{2} + \pi)
  &= \sin(\theta_{1} + \theta_{2} + \pi)
  = -\sin(\theta_{1} + \theta_{2}),
\end{align*}
but $\|f(\theta_{1}, \theta_{2})\| = 1$, so the origin is not in the image of $f$; that is, $f_{1}$ and $f_{2}$ do not vanish simultaneously.
To extend this idea to an $n$-dimensional torus, define
$$
f_{1}(\theta_{1}, \dots, \theta_{n}) = \cos(\theta_{1} + \theta_{2}),\quad
f_{2}(\theta_{1}, \dots, \theta_{n}) = \sin(\theta_{1} + \theta_{2}),\quad
f_{i}(\theta_{1}, \dots, \theta_{n}) = 0,\ i > 2.
$$
The topological point seems to be that a cube is contractible, and "oddness" (even just the sign conditions on the boundary in the Poincaré-Miranda theorem) is enough to imply the image of the boundary of the cube encloses the origin, so by continuity the origin is in the image of $f$. By contrast, an $n$-torus has lots of $1$-dimensional homology, so it's "easy" to wind the image around the origin without hitting the origin, even if the mapping is odd. (At least, that's what naturally suggested the example above.)
