# Showing that the union of three pairwise intersecting sets is not contractible

Let $X$ be a "nice" Hausdorff space (probably a compact complete metric space). Suppose that $A_1,A_2$ and $A_3$ are compact, path-connected subsets of $X$ such that:

• $A_i \cap A_j \ne \varnothing$ is path-connected for all $i,j$; and
• $A_1 \cap A_2 \cap A_3 = \varnothing$.

Is the space $A_1 \cup A_2 \cup A_3$ not contractible, and if so how would you go about showing it?

Also, what are the minimum conditions that can be put on "nice"?

• Presumably you also intend that $A_i \cap A_j$ is nonempty? – Qiaochu Yuan Oct 2 '17 at 23:41
• Intuitively, this seems like an obstruction that the first Čech cohomology group should be able to detect. – Alex Provost Oct 2 '17 at 23:45
• @Robert: the obvious way of interpreting those constructions produces a nonempty triple intersection $A_1 \cap A_2 \cap A_3$, namely the point at which your petals or radii meet. And the obvious way of fixing that problem produces empty double intersections. So can you clarify what you mean? I don't think it is "of course" entirely possible that the union is contractible. – Qiaochu Yuan Oct 2 '17 at 23:55
• @Robert: yes, and a triangle is not contractible. – Qiaochu Yuan Oct 3 '17 at 0:01
• My suggestion would be the following candidate for a noncontractible loop. Let $x_{0}\in A_{1}\cap A_{2}$, $x_{1}\in A_{1}\cap A_{3}$ and $x_{2}\in A_{2}\cap A_{3}$. Compose a path in $A_{1}$ from $x_{0}$ to $x_{1}$, a path in $A_{3}$ from $x_{1}$ to $x_{2}$, and a path in $A_{2}$ from $x_{2}$ to $x_{0}$. That gives you a loop $p$ based at say $x_{0}$. If it is contractible, then at some point in the corresponding path homotopy that image is entirely in $A_{1}\cup A_{2}$. I would try looking there for a contradiction. – Robert Thingum Oct 3 '17 at 0:13

## 1 Answer

It suffices to assume that $X$ is normal (which is stronger than Hausdorff but weaker than metrizable). In particular, any two disjoint closed subsets can be separated by a continuous function (Urysohn's Lemma). Furthermore, you do not need the subsets $A_i$ to be compact, only closed; also, you do not need $A_i\cap A_j$ to be path-connected, only nonempty. I will write a proof as a sequence of lemmas, let me know if you have hard time proving any of these.

Assumption: $X$ is normal, $A_1, A_2, A_3$ are closed, nonempty, path-connected, such that $A_i\cap A_j\ne \emptyset$ for all $1\le i, j\le 3$ and $$\ A_1\cap A_2 \cap A_3=\emptyset.$$

Lemma 1. Given three closed subsets $A_1, A_2, A_3$, as above, $$\exists~ a_i\in A_i - (A_j \cup A_k)$$ for any triple $\{i, j, k\}=\{1, 2, 3\}$.

Lemma 2. There exist open neighborhoods $U_1, U_2, U_3$ of the subsets $A_1, A_2, A_3$ in $X$ such that $$U_1\cap U_2 \cap U_3=\emptyset.$$

Lemma 3. (Urysohn) There exist non-negative continuous functions $\chi_i$, on $X$ such that $\chi_i=1$ on $A_i$ and $\chi_i=0$ on $X- U_i$, $i=1, 2, 3$.

Given these functions, I will define a partition of unity on $$A= A_1\cup A_2 \cup A_3,$$ $$\eta_i= \frac{\chi_i}{\chi_1 + \chi_2 + \chi_3},$$ subordinate to the covering of $A$ by the subsets $U_i\cap A_i$, $i=1, 2, 3$.

The key fact is that the nerve of the above covering is isomorphic to the boundary (topological circle) $S$ of the standard 2-dimensional simplex $$\Delta= \{(t_1, t_2, t_3)\in {\mathbb R}_{\ge 0}^3: t_1+t_2+t_3=1\}.$$ (If you do not know what a "nerve" is, just think of this sentence as defining $S$ and $\Delta$.)

Define the "map-to-nerve" map $$f: A\to \Delta, \quad f(a)= (\eta_1, \eta_2, \eta_3).$$ The image of this map is contained in $S$ and $$f(a_1)= b_1=(1, 0, 0), \ f(a_2)=b_2=(0,1,0),\ f(a_3)=b_3=(0, 0, 1).$$

In addition to the map $f$ we define a map $g: S\to A$ which sends each $b_i$ to $a_i$ and whose restriction to each line segment $[b_i, b_{i+1}]$ ($i$ is taken modulo $3$) is a path in $A_i\cup A_{i+1}$ connecting $a_i$ to $a_{i+1}$. Here I am using path-connectivity of $A_i\cup A_{i+1}$.

Lemma 4. The composition $h:= f\circ g: S\to S$ is homotopic to the identity map. (Hint: Use the fact that the restriction of $h$ to each segment $[b_{i}, b_{i+1}]$ fixes the boundary of this segment and its image is disjoint from $b_{i-1}$.)

We are now ready to prove:

Theorem. $\pi_1(A)\ne \{1\}$, in particular, $A$ is not contractible.

Proof. I claim that the map $g: S\to A$ is not null-homotopic. Suppose to the contrary that this map extends to a continuous map $G: \Delta\to A$. The composition $f\circ G: \Delta\to S$ is homotopic to the identity when restricted to $S$. But this would imply that $\pi_1(S)=1$, which is a contradiction. qed

In fact, you do not need path-connectivity of $A_i$'s, only connectivity. But a proof would require Chech cohomology which you probably are not familiar with.

Edit. Here is a sketch of an argument which assumes mere connectivity of the sets $A_i$, instead of path-connectivity. I will show that $$\check{H}_1(A)\ne 0.$$ One defines Chech homology groups of a topological space $T$ as the inverse limit $$\lim_{\leftarrow} H_1(C_\gamma)$$ where $C_\gamma$'s are nerves of open covers of $T$ where the order is given by the refinement relation of the open covers. The simplicial complexes $C_\gamma(A_i)$ associated with the sets $A_i$ are not just connected but path-connected. Therefore, instead of constructing maps of intervals to $A_i\cup A_j$ as in the above proof, one constructs maps to the nerves of the open covers. As above, one verifies that all the maps $$S\to C_\gamma(A)$$ are not null-homologous. However, more care (in choosing the maps $S\to C_\gamma(A)$) is needed to ensure that these maps define an element of the inverse limit $$\lim_{\leftarrow} H_1(C_\gamma(A)).$$ In the end, you prove that $\check{H}_1(A)\ne 0$. Since the Chech homology is homotopy-invariant, it follows that $A$ is non-contractible.

One can also argue using the Chech fundamental group or the Chech cohomology, whatever is feel more comfortable with.

• Thanks for your response! Assuming that a do know a little Čech cohomology (but am by no means an expert) how would you remove the path-connectivity assumption from the $A_i$'s? I assume you would have to consider refinements of the open cover you've constructed, but I don't see how you would build maps from the nerves of the covers back into $A$. – Zorngo Oct 5 '17 at 0:54
• @Zorngo: I will add an explanation a bit later. – Moishe Kohan Oct 5 '17 at 14:39