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I want to compute $\pi_1(S^1\vee S^2)$, where $\vee$ denotes the wedge sum.

My approach is to eventually apply Van Kampen. (Note that I don't have access to machinery like cell complexes, however, so my approach will be elementary in that regard.)

Also, to be clear, my definition of wedge sum comes from Hatcher: "Given spaces $X$ and $Y$ with chosen points $x_0 \in X$ and $y_0 \in Y$,then the wedge sum $X\vee Y$ is the quotient of the disjoint union $X \sqcup Y$ obtained by identifying $x_0$ and $y_0$ to a single point."

Now, let's set $X=S^1\vee S^2$. And let $z$ be the common point. I want to write $X$ as the union of two open spaces whose intersection is simply connected.

I saw on this post Wedge product $S^1 \vee S^2$ that the OP follows my strategy and outlines the steps. And his answer is confirmed. However, I'm unable to fill in the details due to my own lack of understanding.

I'll outline the steps now and highlight my confusions.

Specifically, we take $x\in S^2$ and $y\in S^1$ where $x, y$, and $z$ are all distinct. Then we set $P=X -\{x\}$ and $Q=X-\{y\}$.

It's now claimed that $X=P\cup Q$. But why? Are we not missing the points $x,y$ from the union?

Then it's claimed that $\pi_1(Q)=\mathbb Z$ because the punctured sphere is homeomorphic to $\mathbb R^2$, which def. retracts onto the point $z$ so that we are just left with $S^1$. I know that the punctured sphere is homeomorphic to $\mathbb R^2$ but I don't see how this fact directly applies or even explicitly how $\mathbb R^2$ would def. retract onto the point $z$. Moreover, I don;t see how this would imply we're left with $S^1$. (I am aware that $\pi_1(S^1)\cong \mathbb Z$.)

Next, it's claimed that $\pi_1(P)$ is trivial because punctured $S^1$ def. retracts to $z$ so that we are left with $S^2$. How?

Finally, it's claimed that $P\cap Q$ is simply connected, but I don't understand this at all either.

As you can tell, likely nothing about the solution to this problem will be obvious to me. I would really appreciate a more detailed explanation of this process.

As a final question, I'd also like to know if $\pi_1(S^1\vee S^2) =\pi_1(A \vee B)$, where $A$ is homeomorphic to $S^1$ and $B$ to $S^2$. I suspect that equality holds, but I'd appreciate an explanation of this as well.

CuriousKid7
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1 Answers1

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1) We have $x \in X \backslash \{y\}$ so $x \in (X \backslash \{y\} \cup \dots)$ and similarly for $y$.

Remark : if $A$ is contractible and meet $B$ in one point $p$, then $\pi_1(A \cup B) \cong \pi_1(B)$. We will use it twice. This is because if $\gamma$ is a path, I can write it as $\gamma_1 \cup \eta_1 \dots \cup \eta_n \cup \gamma_n$ where each $\gamma_i \cap A = p$ and $\eta_i \cap B = p$. Then, $\gamma$ is homotopic to $\gamma_1 \cup \gamma_2 \cup \dots \cup \gamma_n$.

2) $F_t(x,y) = (tx,ty)$ defines a retraction of $\Bbb R^2$ onto the origin. This shows that $S^2 \backslash \{x\}$ retracts on a point. By the remark $\pi_1( S^2 \backslash \{x\} \vee S^1) \cong \Bbb Z$.

3) : $S^1 \backslash \{y\}$ is homeomorphic to an interval, so contractible. Apply again the remark, noticing that $\pi_1(S^2) = 0$.

4) : $P \cap Q$ is simply the wedge of an interval and a plane. They are both contractible and intersect in one point, so you can retract them both at the same time.

5) : If $X \cong Y$ then $\pi_1(X) \cong \pi_1(Y)$. So you just need to show that $A \vee B \cong S^2 \vee S^1$.