# Fundamental group of $S^1\vee S^1$ without applying seifert-van kampen's theorem

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)

• 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 Oct 7, 2020 at 23:29
• @AnthonySaint-Criq It seems nice, but this exercise is before the chapter about covering spaces Oct 7, 2020 at 23:32
• 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. Oct 7, 2020 at 23:33
• How do you prove that $\pi_1(S^1) = \mathbb Z$ without covering spaces? Oct 7, 2020 at 23:33
• @PaulFrost One could resort to the Hurewicz theorem, though I don't recall if the proof relies on covering spaces. Oct 7, 2020 at 23:35

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 $$F_2$$.