Is there a retraction of $S^1\vee S^2$ onto $S^1$?

Question

I am currently working through a problem that requires me to prove that $\pi_1(S^1\vee S^2)\simeq\mathbb{Z}$ directly by showing that the homomorphism induced by the inclusion of $S^1$ on $\pi_1$ is a bijection. I have shown that it is surjective and I now need that it is injective. I know that if there is a retraction of $S^1\vee S^2$ onto $S^1$, then I am done. Is there a retraction of $S^1\vee S^2$ onto $S^1$?

I supposed there was one, and ran through a few of the usual tricks and I didn't get a contradiction. So perhaps there is one. The only thing that I can think of is a map that fixes the circle in the wedge and somehow folds each circle through the north and south pole on the sphere onto the circle in the wedge. But I am not sure if this can be done continuously.

Maybe there is a more straightforward way to show that $i_*$ is injective and I am missing the point. Any thoughts?

Answer 1

The definition of the wedge sum of two pointed spaces $(X,x_0)$ and $(Y,y_0)$ is the quotient space of the disjoint union $X \sqcup Y$ given by identifying the basepoints $x \sim y$. Immediately from the universal properties of the above two constructions (disjoint union and quotient), a continuous map $X \vee Y \to Z$ is given by continuous maps $f_1: X \to Z$ and $f_2: Y \to Z$ such that $f_1(x_0)=f_2(y_0)$.

There is a retract $X \vee Y \to X$ given by $f_1(x)=x$ and $f_2(y)=x_0$. You're just collapsing the second space $Y$ to the basepoint.

Answer 2

Let $x_0\in S^1 \vee S^2$ be the base point. Since $S^1 \vee S^2$ is a coproduct of $S^1$ and $S^2$ and you want to construct a map $r:S^1\vee S^2\rightarrow S^1$, it suffices to determine $r_{\mid S^1}$ and $r_{\mid S^2}$. Just take $r_{\mid S^1}=1_{S^1}$ and $r_{\mid S^2}$ constant map to a point $x_0\in S^1$. Then it will be a retraction.

In the category of pointed spaces there always exists a retraction $r:X\vee Y\rightarrow Y$.

My favourite way of determining fundamental group would be by universal covering. Just note that the universal covering $X$ of that space looks like real line $\mathbb{R}$ with two-dimensional sphere attached to it at every point of $\mathbb{Z}\subseteq \mathbb{R}$ and the covering map: $$\pi:X\rightarrow S^1\vee S^2$$ is given by taking quotient of the obvious $\mathbb{Z}$-action.