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

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. 
