Let $X$ be the unit circle $S^{1}$ with two line segments $\{0\} \times [-1, 1]$ and $[-1,1] \times \{0\}$ as in the figure below. Prove that the fundamental group $\pi_{1}(X, (0,1))$ is non-abelian.enter image description here

I have shown that $S^{1}$ is a retract of $X$, and I considered looking at the sets $U = X\setminus \{a,c\}$ and $V = X \setminus \{b, d\}$ but I’m unsure how to proceed. Any help is appreciated!

  • 1
    $\begingroup$ Can you use the Seifert-Van Kampen theorem? If so you can apply it twice to your situation, first to the upper and lower halves to get their fundamental group and then to the total space. If not, have you tried finding explicit loops which do not commute? $\endgroup$
    – DevVorb
    Commented May 28 at 19:29
  • $\begingroup$ Why do think that the circle is a retract of $X$? $\endgroup$
    – Paul Frost
    Commented May 28 at 20:11
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    $\begingroup$ Since you're talking about retracts, and the question is phrased as "show it's non-abelian" rather than "compute", perhaps an expected approach was to observe that this space retracts onto the subspace "the bottom-left and top-right parts", which is homeomorphic to a wedge of two circles... $\endgroup$ Commented May 29 at 0:20

1 Answer 1


First, notice that we can contract the horizontal line to a point in the middle: this process shows that the figure is homotopically equivalent to the wedge sum of two copies of $S^1$, each one with a vertical line crossing them.

To find the fundamental group of this space, we can use the standard Seifert-Van Kampen argument (similar to the one used to study $S^1\vee S^1$).

Namely, take $U_1$ to be one of the two copies of $S_1$ and half of the other, and define $U_2$ symmetrically.

We have that $U_1\cap U_2$ is homotopically equivalent to a point, so $\pi(U_1\cap U_2)=0$.

Using the same argument as above, we have that $U_1\sim S_1\vee S_1\sim U_2$, so $$\pi(U_1)=\pi(U_2)=\mathbb{Z}*\mathbb{Z},$$ the free group with two generators.

By Seifert-Van Kampen, $\pi(X)=\mathbb{Z}*\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$, the free group with four generators, which is not abelian.

  • $\begingroup$ Just a remark, but it feels mildly more insightful to me to contract the second pair of lines as well, then convince oneself that the result is homotopy equivalent to a wedge of 4 circles. Still finds SVK useful at the end for the wedge's fundamental group. $\endgroup$
    – user176372
    Commented May 29 at 13:49
  • $\begingroup$ @user176372 I agree, but it felt more intuitive to me, geometrically speaking, to only contract the horizontal line. $\endgroup$ Commented May 29 at 14:08

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