Let $X=S^1\vee S^1$ be the wedge sum of two circles. Is there a way to show that $\pi_1(S^1)$ is not isomorphic to $\Bbb Z$, without using the Seifert-van Kampen theorem? (Or is there a way to show that $X$ is not homotopy equivalent to $S^1$?) Since there is a retraction $S^1\vee S^1\to S^1$, and since retraction induces a surjection on $\pi_1$, I see that $\pi_1(X)$ is inifnite. But I can't show $\pi_1(X)$ is not $\Bbb Z$. (I need this to solve an exercise (Hatcher, 1.1.16 (e)) in algebraic topology, but it is in a chapter before Seifert-van Kampen theorem)

  • 1
    $\begingroup$ What about computing the automorphisms of the universal cover like for the circle ? Prove that there are two infinite order automorphisms that do not commute, which is not so hard to picture $\endgroup$ Oct 7, 2020 at 23:29
  • $\begingroup$ @AnthonySaint-Criq It seems nice, but this exercise is before the chapter about covering spaces $\endgroup$
    – blancket
    Oct 7, 2020 at 23:32
  • 2
    $\begingroup$ Why don't you just tell us the exercise? You may not need this result to solve it, especially if you don't have either covering spaces or Seifert-van Kampen available. $\endgroup$ Oct 7, 2020 at 23:33
  • $\begingroup$ How do you prove that $\pi_1(S^1) = \mathbb Z$ without covering spaces? $\endgroup$
    – Paul Frost
    Oct 7, 2020 at 23:33
  • $\begingroup$ @PaulFrost One could resort to the Hurewicz theorem, though I don't recall if the proof relies on covering spaces. $\endgroup$
    – Aweygan
    Oct 7, 2020 at 23:35

1 Answer 1


There are two retractions $p_1, p_2 : S^1 \vee S^1 \to S^1$ which induce two different surjections $\pi_1(S^1 \vee S^1) \to \pi_1(S^1) \cong \mathbb{Z}$, and in fact together they give a map $S^1 \vee S^1 \to S^1 \times S^1$ which induces a surjection on $\pi_1$ (you can show this very explicitly by just lifting a pair of loops to a loop). So $\pi_1(S^1 \vee S^1)$ admits a surjection onto $\mathbb{Z}^2$ and hence can't be $\mathbb{Z}$.

Exercise 1.1.16(e) in Hatcher asks you to show that a disk with two boundary points identified $D$ doesn't retract onto its boundary $S^1 \vee S^1$. You can do this more directly than the above, by just showing that the induced map $\pi_1(S^1 \vee S^1) \to \pi_1(D) \cong \mathbb{Z}$ isn't injective. And this is pretty straightforward: the two loops in $S^1 \vee S^1$ map to the same element of $\pi_1(D)$ but they can be distinguished by either of the retractions $p_1, p_2$. (This doesn't even require that you be able to calculate $\pi_1(D)$.)

Without Seifert-van Kampen but with some other tools more options are available if you just want to show that $\pi_1(S^1 \vee S^1)$ can't be $\mathbb{Z}$. If homology is available you can compute that $H_1(S^1 \vee S^1) \cong \mathbb{Z}^2$. And if covering spaces are available it suffices to exhibit a Galois cover with non-cyclic Galois group, or show that there are non-isomorphic covers of the same degree, and it's easy to write these down explicitly. In fact you can classify the covering spaces by hand in a way that will prove that the fundamental group is $F_2$ without Seifert-van Kampen, e.g. by writing down the universal cover as the Cayley graph of $F_2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.