A common example of a weak homotopy equivalence which isn't symmetric is the pseudo circle $\mathbb{S}$. Wikipedia gives the following map $f\colon S^{1}\rightarrow\mathbb{S}$ $$f(x,y)=\begin{cases}a\quad x<0\\b\quad x>0\\c\quad(x,y)=(0,1)\\d\quad(x,y)=(0,-1)\end{cases}$$

As to why it's a weak homotopy equivalence, Wikipedia says

This can be proved using the following observation. Like $S^1$, $\mathbb{S}$ is the union of two contractible open sets $\{a,b,c\}$ and $\{a,b,d\}$ whose intersection $\{a,b\}$ is also the union of two disjoint contractible open sets $\{a\}$ and $\{b\}$.

I understand why $f$ is continuous and why the quote is true. I have no idea why the quote shows that the fundamental group of the pseudo circle is infinite cyclic, other homotopy groups are zero, and $f$ induces an isomorphism on fundamental groups. Would you please tell me why?

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2 Answers2


The observations allow you to argue much in the same way you do for $S^1$. For the case of the fundamental group, you can construct a "universal cover" $X$ for $\mathbb{S}$ called the digital line. The space $X$ can be viewed as the set of integers $\mathbb{Z}$ with topology generated by the sets $\{2n\}$ and $\{2n,2n+1,2n+2\}$. The "covering map" $p \colon X \to \mathbb{S}$ sends even numbers to either $a$ or $b$ and odd numbers to either $c$ or $d$. (You should be able to figure out the specifics here. We want $X$ to wrap around $\mathbb{S}$, much like we have $\mathbb{R}$ wrapping around $S^1$.) The usual argument for computing the fundamental group of $S^1$ applied in this setting.

The case of higher homotopy groups is similar.

Once you have that, then it suffices to check that your map $f$ above sends a generator of $\pi_1(S^1)$ to a generator of $\pi_1(\mathbb{S})$.

Balarka Sen
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  • I refer people to Grothendieck's comments in Esquisse d'un programme, section 2, – Ronnie Brown Jul 07 '19 at 20:47
  • I refer people to Grothendieck's comments in Esquisse d'un programme, section 2, quoted in English translation in https://mathoverflow.net/questions/220561/descent-theorems-for-fundamental-groups-and-groupoids. The use here of groupoids was also suggestive of higher dimensional uses of strict homotopy groupoids. – Ronnie Brown Jul 07 '19 at 20:56

If a space $X$ is the union of two path connected open sets $U,V$ whose intersection $W=U \cap V$ has $n$ path components, then the natural thing is to choose a set $A$ consisting of one point in each path component of $W$; the form of the Seifert-van Kampen Theorem given in Topology and Groupoids, Section 6.7, determines the fundamental groupoid $\pi_1(X,A)$ as a pushout of groupoids, and from this one needs a bit of "combinatorial groupoid theory" to determine the various fundamental groups. Thus if $U,V$ are contractible, then the fundamental groups are free groups on $(n-1)$ generators.

What set me on the "groupoid path" in 1965-1968 was finding that it seemed that all of $1$-dimensional homotopy theory was quite naturally expressed in terms of groupoids rather than groups, yielding more powerful theorems with in some cases simpler proofs. In the case in point, there is no need for a diversion to covering spaces.

19 July, 2015: Here is a link to a small correction to an exposition of T&G on the Jordan Curve Theorem. It uses the algebra of groupoids to show that if $$\begin{matrix} C & \xrightarrow{b} & B \\ a \downarrow &&\downarrow\\ A & \to & G \end{matrix}$$ is a pushout of groupoids such that $C$ is totally disconnected, i.e. is a disjoint union of groups, $G$ is connected, and $a,b$ are bijective on objects, then $G$ contains a free groupoid as a retract.

T&G is currently the only topology text in English to give this form of the van Kampen Theorem for nonconnected spaces. It is also given in this downloadable book Categories and Groupoids.

See also this mathoverflow discussion on "is there compelling evidence that two base points are better than one?".

Of course the covering space argument is useful to show the higher homotopy groups of the pseudocircle are trivial.

Ronnie Brown
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