Fill a blank - Algebraic Topology

I am trying to solve the following problem in Allen Hatcher's Algebraic Topology book :

• Let $A_1, A_2, A_3$ be compact sets in $\mathbb R^3$. Use the Borsuk-Ulam theorem to show that there is one plane $P \subset \mathbb R^3$ that simultaneously divides each $A_i$ into two pieces of equal measure.

I managed to find the good idea, but I am stuck with this detail that I cannot find a way to explain. At some point in my proof, I am given a continuous function $f : S^2 \times [a,b] \to [0,1]$ such that for every $v \in S^2$, there exists $t \in [a,b]$ with $f(v,t) = 1/2$. Furthermore, I know that the map $f(v,-) : [a,b] \to [0,1]$ is decreasing and surjective (maybe this is not necessary, but it is how I have shown that $t$ with $f(v,t) = 1/2$ existed...).

I want to show the existence of a continuous map $\varphi : S^2 \to S^2 \times [a,b]$ such that $f \circ \varphi$ is the constant map at $1/2$ (I expect $\varphi(v) = (v,t(v))$ for some function $t : S^2 \to [a,b]$). If I can do this then my proof is complete and using Borsuk-Ulam's theorem I have proved the statement.

I must add that I expected $t(v)$ to be as follows : if $f(v,-)$ is such that $f(v,-)^{-1}(\{1/2\}) = [t_{min}(v),t_{max}(v)]$, I thought $$t(v) = \frac{t_{min}(v) + t_{max}(v)}2.$$ This choice of $t(v)$ has the more elegant property that $f(-v,t(-v)) = f(v,-t(v))$ because of the way $f$ is defined, but I don't know how to show it will be continuous, even though I expect it to be.

Any ideas?

Here is an idea: the space $S^2 \times [a, b]$ can be thought of as a solid spherical shell of thickness $[a,b]$. Take your function $f:S^2 \times [a, b] \to [0,1]$, and define a function $g : S^2 \to S^2 \times [a, b]$ in the following way: for a given $v \in S^2$, $g(v) = (v, t_v)$ where $t_v$ is the smallest value of $t$ with $f(v, t_v) = 1/2$. Let us put $E=g(S^2)$. By construction, $E$ contains exactly one point in each fibre of the projection $S^2\times [a,b] \to [a, b]$. It is like a "slice" of the solid spherical shell $S^2 \times [a, b]$. The projection $S^2 \times [a, b] \to S^2$ restricts to a map $E \to S^2$. This map is an isomorphism with inverse $g$. By construction, the function $f$ is identically $1/2$ on $E$, so the inclusion $E \to S^2\times [a,b]$ is the map $\varphi$ you are looking for (once we identify $E$ with $S^2$).
• Yes, that I figured! But why should $g$ be continuous? That is precisely where I am stuck and it is why I am asking this on MSE. – Patrick Da Silva Sep 24 '13 at 16:55
• @PatrickDaSilva It was far from clear from your post that you had gotten this far. ;) Continuity of $g$ should be a consequence of the fact that $f(v, -)$ is decreasing. I'll let you figure it out! – Bruno Joyal Sep 24 '13 at 16:59
• I agree that it wasn't clear from my question, I guess I should've explained it. But I really can't figure out why $g$ should be continuous... – Patrick Da Silva Sep 24 '13 at 17:00