Topology Question on a Particular Surface I have a following question that states:

Let $T$ be (the surface of) my morning iced doughnut (a torus, not the jam-filled kind!).  For the purpose of this problem we shall assume that $T$ is symmetric with respect to a coordinate system whose origin is in the middle of the hole, so that if $x\in T$, then $-x\in T$ too.  Let $f(x) $ denote the depth of the icing at a point $x$ on $T$, a function which we shall assume to be continuous.
Prove that these is a point $x$ on my doughnut at which $f(x) = f(-x)$.

(original scan)
I am quite unsure of how to progress. My first problem is understanding the covering of the icing. Conventionally, icing only covers a particular part of the donut. Should I assume that? Am I working in 3 dimensions? That is, is my function one that $\mathbb{R^3} \mapsto \mathbb{R}$? This question is one that has been provided in my topology class, but I can't seem to bring topological concepts to light here. Is this torus homeomorphic to something that I can work with? 
Any help would be greatly appreciated. Thank you.
 A: The intermediate value theorem suffices: Consider the function $g(x):=f(x)-f(-x)$ on $T$, which is continuous. If $g(x_0)\ne0$ at some point $x_0\in T$, say $g(x_0)>0$, then $g(-x_0)<0$. Since the surface $T$ is connected there is a continuous path $$\gamma: \quad [0,1]\to T,\qquad t\mapsto\gamma(t),$$ with $\gamma(0)=-x_0$, $\>\gamma(1)=x_0$. The pullback $$\phi(t):=g\bigl(\gamma(t)\bigr)$$ is continuous and satisfies 
$$\phi(0)=g(-x_0)<0,\quad \phi(1)=g(x_0)>0\ .$$Therefore $\phi(\tau)=0$ for some $\tau$ in the interval $\ ]0,1[\ $. Let $\xi:=\gamma(\tau)\in T$. As $g(\xi)=\phi(\tau)=0$ we have  $f(\xi)=f(-\xi)$.
A: The icing covers the entire donut, but if the depth of icing $f(x)$ is zero at some point $x$, you might imagine that there is no icing at $x$; in the context of the problem you're solving, it doesn't matter.
The function is $f:T\to [0,\infty)$.  It is defined only at each point of the surface of the donut (that is $T$, not all of $R^3$) and its value is a non-negative icing depth, which is a number that is at least $0$.
I think it might suffice to consider the largest circle around the outside of the donut, so you are ignoring all of $T$ except for a subset which is equal to $S^1$ and which is symmetric with respect to the origin.  Then I think the usual way to proceed is to let $g(x) = f(x)-f(-x)$ and show that there must be some $x$ for which $g(x)=0$.  Have you studied any fixed-point theorems?  These seem to be to be relevant here.
