Asphericity of pseudo circle

algebraic-topologygeneral-topologyhomotopy-theory

Let $S$ be de pseudo circle (see for instance https://en.wikipedia.org/wiki/Pseudocircle).

How can I compute its higher homotopy groups?

The question came up while trying construct a weak homotopy equivalence $X\to Y$ that does not admit an inverse. Taking $X=S^1$, $Y=S$ and $f$ as in the wikipedia page should work, but I was not able to work out a "simple" way of calculating $\pi_n(S)$ for $n>1$. By "simple" I mean that it should not invoke a generalization of that I am trying to prove.

Best Answer

The most elementary way I know of is to construct its universal cover. First, some preliminaries. We consider any preordered set as a topological space where the open sets are the downward closed sets; a map between two preordered sets is then continuous iff it is order-preserving. The following Lemma is very convenient for thinking about homotopies of such maps:

Lemma: Let $P$ and $Q$ be preordered sets and $f,g:P\to Q$ be order-preserving maps such that $f(x)\leq g(x)$ for all $x\in P$. Then $f$ and $g$ are homotopic.

Proof: Define $H:P\times [0,1]\to Q$ by $H(x,t)=f(x)$ for $t\in [0,1)$ and $H(x,1)=g(x)$. If $U\subseteq Q$ is downward closed, then $$H^{-1}(U)=f^{-1}(U)\times[0,1)\cup g^{-1}(U)\times\{1\}=f^{-1}(U)\times[0,1)\cup g^{-1}(U)\times[0,1]$$ since $g^{-1}(U)\subseteq f^{-1}(U)$. Since $f^{-1}(U)\times[0,1)\cup g^{-1}(U)\times[0,1]$ is clearly open in $P\times[0,1]$, this shows $H$ is continuous.

Corollary: Let $P$ be a preordered set with an element $p$ which is comparable to exactly one other element $q$ of $P$. Then $P$ deformation-retracts onto $P\setminus\{p\}$.

Proof: Define $f:P\to P$ by $f(p)=q$ and $f(x)=x$ for $x\neq p$. Then $f$ is order-preserving and either $f\leq 1_P$ or $f\geq 1_P$ depending on whether $q\leq p$ or $q\geq p$. Either way, by the Lemma, $f$ is homotopic to $1_P$ and this gives a deformation retraction from $P$ to $P\setminus\{p\}$.

Let's now turn to the pseudocircle $S$. We write $S=\{a,b,c,d\}$ with $a,b\leq c,d$.

The universal cover can then be described as $S\times\mathbb{Z}$, with the order given by $(a,n)\leq (d,n-1),(c,n)$ and $(b,n)\leq (c,n),(d,n)$. (Draw a picture of this poset! It should look like a zigzag path with $a$'s and $b$'s alternating on bottom and $c$'s and $d$'s alternating on top.) The covering map $p:S\times\mathbb{Z}\to S$ is then just the projection; it is easy to verify that it is indeed a covering map.

I now claim that $S\times\mathbb{Z}$ is weakly contractible, so that the higher homotopy groups of $S$ are trivial. Every compact subset of $S\times\mathbb{Z}$ is finite, so it suffices to show that $S$ is covered by an increasing sequence of contractible finite subsets. To prove this, note first that $\{(a,0),(c,0)\}$ deformation-retracts to $\{(a,0)\}$ by the Corollary since $(a,0)\leq (c,0)$. Similarly $\{(a,0),(c,0),(b,0)\}$ deformation-retracts to $\{(a,0)(c,0)\}$ since $(b,0)\leq (c,0)$ and $(b,0)$ is not comparable to $(a,0)$. In the same way $\{(a,0),(c,0),(b,0),(d,0)\}$ deformation-retracts to $\{(a,0),(c,0),(b,0)\}$ since $(d,0)$ is only comparable to $(b,0)$. We can continue adding points one-by-one in this way: $(a,1)$ is only comparable to $(d,0)$, then $(c,1)$ is only comparable to $(a,1)$, and so on. We can also add points "in the other direction": $(d,-1)$ is only comparable to $(a,0)$, and then $(b,-1)$ is only comparable to $(d,-1)$, and so on. So, we can add on all of the points of $S\times\mathbb{Z}$ one by one, maintaining a contractible space at each step.

(In fact, with a bit more work, you can show it's possible to glue together all these deformations in sequence continuously to show that $S\times\mathbb{Z}$ is contractible, not just weakly contractible.)

Alternatively, using a bit of heavier machinery from homotopy theory, you can show quickly and directly that there is a weak homotopy equivalence $f:S^1\to S$. Specifically, pick two points $p,q\in S^1$ and let $f$ map $p$ to $c$, $q$ to $d$, and the two open arcs between $p$ and $q$ to $a$ and $b$ respectively. Note that $S$ is covered by the open sets $U=\{a,b,c\}$ and $V=\{a,b,d\}$ which are both contractible (they deformation-retract onto $\{c\}$ and $\{d\}$ respectively, either using the Corollary twice or directly using the Lemma). Also, $f^{-1}(U)=S^1\setminus\{q\}$ and $f^{-1}(V)=S^1\setminus\{p\}$ are contractible, so $f$ restricts to homotopy equivalences $f^{-1}(U)\to U$ and $f^{-1}(V)\to V$. Moreover, $f$ also restricts to a homotopy equivalence $f^{-1}(U\cap V)\to U\cap V$, since $U\cap V=\{a,b\}$ is discrete and $U\cap V$ is a disjoint union of two arcs, one mapping to each of $a$ and $b$. It follows by the theorem below that $f:S^1\to S$ is a weak homotopy equivalence.

Theorem: Let $f:X\to Y$ be a continuous map and let $C$ be an open cover of $Y$ such that for each nonempty finite $F\subseteq C$, $f$ restricts to a weak homotopy equivalence $f^{-1}(U_F)\to U_F$, where $U_F$ is the intersection of all the elements of $F$. Then $f$ is a weak homotopy equivalence.

You can find a proof of (a slight strengthening of) this theorem as Theorem 6.7.11 in tom Dieck's Algebraic Topology, for instance.

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