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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)

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    $\begingroup$ 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
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    $\begingroup$ 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

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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 $F_2$.

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