Retraction between wedge sum and product of spaces Is there a retraction between $S^1 \vee \mathbb{D}^2 $ and $S^1 \vee S^1 $ or between $S^1 \times \mathbb{D}^2  $ and $S^1 \times S^1 $?
I know that for a retraction to exist between $X$ and $A$ there must be a surjective homomorphism from $\pi _1(X)$ to $\pi _1 (A)$. SO for the first case there must be a surjection from $\mathbb{Z}$ to $F_2 $ (free group on two generators) and for the second pair there must exist a surjective homomorphism from $\mathbb{Z}$ to $\mathbb{Z} \times \mathbb{Z} $ and can't rule out whether these homomorphisms could exist or not.
Perhaps the fact that these homomorphisms would be completely determined by the value of $\phi (1) $ (where $\phi $ is our homomorphism from the integers) since they're homomorphisms from $\mathbb{Z}$. Because of this I'm inclined to say I don't think these homomorpisms exist, but don't know how to show it.
 A: Let's start by showing there's no surjection from $\mathbb{Z}$ to $F_2$.
In fact, there is no surjection from any Abelian group to any non-Abelian. For if $f: G \to H$ is a surjective group homomorphism and $G$ is Abelian, then for every $a, b \in H$, we could take some $x, y$ such that $f(x) = a$, $f(y) = b$ and show that $ab = f(x) f(y) = f(xy) = f(yx) = f(y) f(x) = ba$. Since $F_2$ is not Abelian and $\mathbb{Z}$ is Abelian, there can be no group homomorphism surjection $\mathbb{Z} \to F_2$.
Now, we show there is no group homomorphism surjection $f : \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$. For note that the only group homomorphisms possible are those of the form $f(x) = (x a, xb)$. And none of these surjections can possibly be a surjection. For if $a = 0$, then there is no $x$ such that $f(x) = (1, 1)$. And if $a \neq 0$, then there is no $x$ such that $f(x) = (0, 1)$.
In fact, once we have that there's no group homomorphism surjection $\mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$, we note that there can't be a homomorphism $\mathbb{Z} \to F_2$ since if there were, we could compose it with the surjective homomorphism $F_2 \to \mathbb{Z} \times \mathbb{Z}$ sending $a$ to $(1, 0)$ and $b$ to $(0, 1)$. So the first proof was not necessary.
