I am trying to compute $\pi_1(S^1 \vee S^2$) by Van Kampen. I know Hatcher has a solution but I need to verify if my approach is correct and rigorous. I have seen a previous post on this topic, but I am using a different decomposition of $X=S^1 \vee S^2$, so please bear with me.

Let $z$ be the common point (wedge point). Let $x_0$ be another point on $S^2$, and $x_1$ be another point on $S^1$.

Then let $Q=X\backslash x_0$, and $P=X\backslash x_1$.

Clearly, $X=Q\cup P$, and $Q,P$ both open.

Now, $\pi_1(Q)=\mathbb Z$, since the punctured sphere is homeomorphic to $R^2$, which def. retracts to the point $z$ and we are left with just $S^1$.

Similarly, $\pi_1(P)$ is trivial, since punctured $S^1$ def. retracts to $z$, and we are left with $S^2$, which is simply connected.

Also, $\pi_1(P\cap Q)$ is wedge of punctured sphere with punctured circle, both of which def. retract to $z$, and hence the wedge is simply connected.

So, now by Van Kampen, we obtain that $\pi_1(P\cup Q)$ is isomorphic to $\pi_1(P)_{*\pi_1(P\cap Q)}\pi_1(Q)$, which is just $\mathbb Z$.

Is this proof rigrous? Have I made some assumptions that have not been justified?

  • 4
    $\begingroup$ This is right. $ $ $\endgroup$
    – Potato
    Commented Jun 3, 2013 at 21:28

1 Answer 1


It's correct, even intuition suggest us that in this case the only kind of paths that aren't in the same equivalence class of the trivial one is the paths around $S^1$. And moreover it is not important that they move around the sphere because I can homotopically retract them to move only around the circle if they move around it.

if fact you have just applied a more general corollary of Van Kampfen which is

$\pi_1(X \vee Y) $ is the group with generators the union of generators of $\pi_1(X)$ and $\pi_1(Y)$ and with relations the union of relations of the two groups

the proof is easy, just choose the two opens $X$, $Y$ and intersection a simply connected neighborhood of the wedge point

  • $\begingroup$ Thanks, intuitively it was clear to me (and I know the answer is correct since Hatcher says so). I just wanted to write down all steps rigorously and make sure I didn't miss anything. $\endgroup$ Commented Jun 3, 2013 at 21:45
  • 1
    $\begingroup$ You're welcome, (I prefer using the notation with generators and relations instead of quotient of groups but it's the same. Obviously. :) $\endgroup$
    – Riccardo
    Commented Jun 3, 2013 at 21:53
  • 2
    $\begingroup$ Naturally you want that $X$ and $Y$ satisfy some minimal regularity hypotheses, which in this case appear to be "locally simply connected". $\endgroup$
    – Ryan Reich
    Commented Jun 3, 2013 at 22:14
  • $\begingroup$ Surely, my apologies to not mention it. I was thinking about the example given by the OP that satisfy this request. Thank you for the observation $\endgroup$
    – Riccardo
    Commented Jun 3, 2013 at 22:28

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