I'm having some difficulty getting an understanding of this issue: I have an inclusion map $i : S^1 \vee S^1 \hookrightarrow S^1 \times S^1$. So this is an inclusion map from the figure eight to the torus. If the leftmost circle of the figure eight is called $b$ and the rightmost circle is called $a$, then $i(b)$ is the circle which comprises the center of the torus and $i(a)$ corresponds to a longitudinal circle of the torus.
I am considering four questions pertaining to this inclusion map.
The first is, Is $i_*: \pi_1(S^1 \vee S^1) \to \pi_1(S^1 \times S^1)$ injective? My intuition is that no, this is not injective because $\pi_1(S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z}$, the free group on two generators and $\pi_1 (S^1 \times S^1) = \mathbb{Z}\times \mathbb{Z}$. However, I am not sure if this is in fact true and I am trying to figure out the best way to go about showing it.
Is $i_*$ surjective? My intuition is that it is not surjective, but I am not confidentin in my intuition on this. What would be the best way to try to prove or disprove the surjectivity of $i_*$?
Does $i$, the original inclusion map from the figure eight to the torus, have a retraction back? I believe that it does not. My understanding is that a retraction exists from the punctured torus to the figure eight, but I do not believe that there is a retraction from $S^1 \times S^1$ to $S^1 \vee S^1$. Is this understanding correct? What would be the best way to demonstrate it?
Finally, is $i$ a homotopy equivalence? My intuition is that $i$ is not a homotopy equivalence because I am having trouble visualizing a continuous deformation of the torus into the figure and vice versa. The torus has one whole and the figure eight has two. However, I do not know if my intuition is correct here. How would one show this?
Thank you for any and all assistance!