Erratum to: Groupoids, the Phragmen–Brouwer property, and the Jordan curve theorem

1 Erratum to: J Homotopy Relat Struct (2006) 1:175–183

Abstract   Omar Antolín Camarena pointed out a gap in the proofs in Brown (Topology and Groupoids. Booksurge LLC, South Carolina 2006; J Homotopy Relat Struct 1:175–183, 2006) of a condition for the Phragmen–Brouwer property not to hold; this note gives the correction in terms of a result on a pushout of groupoids, and some additional background.

2 Introduction

This note fills in two ways a gap in a proof in [2], as explained in Sect. 2.

The paper [2] shows how groupoid methods can be used to relate fundamental groups to a classic separation property of spaces, often called the Phragmen–Brouwer property. However it turns out, as we explain in Sect. 2, that the applications to the Jordan Curve Theorem require the following stronger result on groupoids:

Theorem 1.1

Suppose given a pushout of groupoids

(1)

such that $i,j$ are bijective on objects, $C$ is totally disconnected, and $G$ is connected. Then $G$ contains as a retract a free groupoid whose vertex groups are of rank

$$\begin{aligned} k=n_C-n_A-n_B+1, \end{aligned}$$

where $n_P$ is the number of components of the groupoid $P$ for $P=A,B,C$ $($assuming these numbers are finite$).$

Further$,$ if $C$ contains distinct objects $a,b$ such that $A(ia,ib),B(ja,jb)$ are nonempty, then $F$ has rank at least $1.$

The proof is given in Sect. 3. It is easily seen how the theorem applies to topological situations like the following:

for which covering space methods are more awkward.

3 The Phragmen–Brouwer property

One source of confusion is that there are two forms of this property, as follows.

Let $X$ be a topological space. We say a subspace $D$ of $X$ separates $X$ if $X{\setminus } D$ has more than one component. We say a subset $D$ of $X$ separates the points a and b of X if $a$ and $b$ lie in distinct components of $X {\setminus } D$.

A topological space $X$ is said to have the Phragmen-Brouwer Property I, here abbreviated to (PBI), if $X$ is connected and the following holds:

(PBI): If $D$ and $E$ are disjoint closed subsets of $X$ and $a, b$ are points of $X$ not in $D\cup E$ such that neither $D$ nor $E$ separate $a$ and $b$, then $D \cup E$ does not separate $a$ and $b$.

We also consider:

(PBI$'$): If $D$, $E$ are disjoint closed subsets of $X$ such that neither $D$ nor $E$ separate $X$, then $D \cup E$ does not separate $X$.

Such a space $X$ is also called unicoherent. In [11, pp. 47–49] it is proved that these two properties (PBI), (PBI$'$), are equivalent if $X$ is connected and locally connected.

It is of interest and importance to relate these properties to the fundamental group of the space $X$; see [5, Section 3], and for further information on this area see [6, 11].

Now we can state what is the gap in [2]: Proposition 4.1 in that paper shows that if $X$ is connected and does not satisfy (PBI$'$), then the fundamental group of $X$ contains the integers as a retract. However what is needed for the Jordan Curve Theorem is actually Corollary 3.1 in the next section, which is the same criterion but for $X$ not satisfying (PBI). This is required because the proof of Proposition 4.3 in [2] uses (PBI). The equivalence of (PBI) and (PBI$'$) fills the gap, but the proof in the next section fits better with the goal of the original proof of letting the algebra of groupoids do most of the work and minimizing the point-set topology input.

4 Proof of Theorem 1.1

Let $\mathsf {Fr}\,:{\mathsf {DirectedGraphs}} \rightarrow {\mathsf {Groupoids}}$ be the free groupoid functor. From [3, Chap. 8]¹ it follows that any groupoid $G$ has a retraction $G \rightarrow \mathsf {Fr}\,W$ where $W$ is a forest.

Let $Z$ be the set of objects of $C$ (and of $A, B$ and $G$) regarded as a directed graph with no edges. Pick spanning forests $X$ and $Y$ of the underlying directed graphs of $A$ and $B$. Then there are retractions

$$\begin{aligned} C\rightarrow \mathsf {Fr}\,Z,\quad A \rightarrow \mathsf {Fr}\,X,\quad B \rightarrow \mathsf {Fr}\,Y. \end{aligned}$$

By “span” in a category we mean a pair of arrows $U \leftarrow W \rightarrow V$; this is the shape of a diagram whose pushout you can take. Then the following diagram in ${\mathsf {Groupoids}}$ commutes and its rows are spans:

So the span $\mathsf {Fr}\,X \leftarrow \mathsf {Fr}\,Z \rightarrow \mathsf {Fr}\,Y$ is a retract of the span $A \leftarrow C \rightarrow B$. This implies that the pushout, say $F$, of $\mathsf {Fr}\,X \leftarrow \mathsf {Fr}\,Z \rightarrow \mathsf {Fr}\,Y$ is a retract of $G$ (which is the pushout of $A \leftarrow C \rightarrow B$). Since the span of free groupoids is actually the image under $\mathsf {Fr}\,$ of the obvious span $X \leftarrow Z \rightarrow Y$ of graphs, and since $\mathsf {Fr}\,$ is a left adjoint, this pushout $F$ is actually just $\mathsf {Fr}\,W$ where $W$ is the pushout in the category of directed graphs of $X \leftarrow Z \rightarrow Y$.

This graph $W$ is connected because $G$ is connected, so, denoting by $e$(Q) and $v$(Q) the number of vertices of a graph, the vertex groups in $\mathsf {Fr}\,W$ are free of rank $k = e(W) - v(W) + 1$. We have $v(W) = v(X) = v(Y) = v(Z) = n_C$; and, since $Z$ has no edges, $e(W) = e(X) + e(Y)$. Also, since $X$ is a spanning forrest we have $e(X) = v(X) - n_A = n_C - n_A$, and similarly, $e(Y) = n_C - n_A$. Putting this all together, the vertex groups in $F$ have rank $(n_C - n_A) + (n_C - n_B) - n_C + 1 = n_C - n_A - n_B + 1$, as claimed.

For the last part of the theorem, we choose $X, Y$ so that the elements $\alpha , \beta$ of $A(ia,ib), B(jb,ja)$ respectively are parts of $\mathsf {Fr}\,X, \mathsf {Fr}\,Y$ respectively. These elements map to elements $\alpha ', \beta '$ in $F$ and the element $\alpha '\beta '$ will be nontrivial in $F$; so $F$ has rank at least $1$. This completes the proof.

The next corollary is an essential part of the proof of the Jordan Curve Theorem. It appears, without the retraction condition, as part of [10, Theorem 63.1], and also as [2, Proposition 4.1], [3, 9.2.1].

Corollary 3.1

If the space $X$ is path connected and does not have the $($PBI$),$ then its fundamental group at any point contains the infinite cyclic group as a retract.

Proof

The proof now follows the methods of [2], replacing that paper’s Corollary 3.5 with Theorem 1.1. $\square$