Does there exist a continuous partition of the sphere into sets of cardinality 4?

Define $X^{\{n\}}:=\{A\subseteq X:|A|=n\}$, the set of subsets of cardinality $n$. If $X$ is a topological space, $X^{\{n\}}$ can be given a topology by considering it to be a quotient of $X^n$ minus the extended diagonal.

Define a continuous $n$-partition of a space $X$ to be a partition of $X$ into sets of cardinality $n$ such that the associated function $X\to X^{\{n\}}$, sending $x$ to the element of the partition that contains it, is continuous.

(The notation and terminology is my own. I don't know if there is standard notation or terminology for these.)

Does there exist a continuous $4$-partition of the sphere?

My assumption is no, because most obvious $4$-partitions fail. If we could find a $2$-partition (also known as a fixed-point-free involution) of the projective plane ${\rm\Bbb RP}^2$, we'd be golden, since each element of ${\rm\Bbb RP}^2$ corresponds to an antipodal pair of the sphere $S^2$, but this is impossible by the Lefschetz fixed point theorem.

There's probably some simple algebraic topology solution for this, but unfortunately, I am unskilled in algebraic topology.

For a further conjecture, I believe $S^2$ has size $2$ mod $4$, in the following sense: if $k$ points are deleted from $S^2$, then the result has a continuous $4$-partition iff $k\equiv2 \pmod{4}$.

Solution 1:

Let $X$ be a manifold and suppose $f:X\to X^{\{n\}}$ is a continuous $n$-partition. Let $p:X\to Y$ be the quotient map associated to this partition of $X$. I claim that in fact $p$ is a covering map. (Conversely, it is easy to see that any partition that comes from an $n$-sheeted covering map is continuous.)

To prove this, let $y\in Y$, so $y=\{x_1,\dots,x_n\}$ is an $n$-element subset of $X$. Let $U_1,\dots,U_n$ be pairwise disjoint neighborhoods of $x_1,\dots,x_n$. Then there is an open subset $U$ of $X^{\{n\}}$, consisting of sets which contain exactly one element from each $U_i$, and the topology of $U$ is just the product topology when you identify it with $\prod U_i$. By continuity of $f$, there is an open neighborhood $V_1\subseteq U_1$ of $x_1$ such that $f(V_1)\subseteq U$. Restricting $f$ to a map $V_1\to U\cong \prod U_i$, we can think of it as a collection of continuous maps $f_i:V_1\to U_i$ for each $i$. Since $f$ is a partition, each $f_i$ is injective. By invariance of domain (this is the one place we are using the assumption that $X$ is a manifold), this means $f_i(V_1)$ is open for each $i$ and $f_i$ is a homeomorphism $V_1\to f_i(V_1)$. Let us write $V_i=f_i(V_1)$.

So, we have open neighborhoods $V_1,\dots,V_n$ of $x_1,\dots,x_n$ and homeomorphisms $f_i:V_1\to V_i$ such that our partition just splits $\bigcup V_i$ into the sets of the form $\{f_1(x),f_2(x),\dots,f_n(x)\}$ for each $x\in V_1$. So, the set $W=p(V_1)=p(V_2)=\dots=p(V_n)$ is an open neighborhood of $p$ in $Y$, and it is evenly covered by $p$, with $p^{-1}(W)=\bigcup V_i$ and $p$ mapping each $V_i$ to $W$ homeomorphically.

In particular, in the case of $S^2$, a continuous $n$-partition would give a surface that has $S^2$ as an $n$-sheeted covering space. By the classification of surfaces, this is possible for only $n=1,2$.

Furthermore, if you have a continuous $n$-partition of an $k$-times punctured sphere $X$, this gives an $n$-sheeted covering map $p:X\to Y$. The space $Y$ must then be a connected surface with finitely generated fundamental group, and thus is a surface of finite type (you can show it has finitely many ends and its end compactification is a closed surface). It follows that $\chi(X)=n\chi(Y)$. Since $\chi(X)=2-k$, this means $k$ must be $2$ mod $n$.