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.
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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.
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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 $i_*$?
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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?
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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!
Best Answer
Regarding 1:
Consider a homomorphism $f: \mathbb{Z} \ast \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$. If $a$ and $b$ are the generators of $\mathbb{Z} \ast \mathbb{Z}$ then consider what $ab$ and $ba$ map to:
$f(ab) = f(a)f(b) = f(b) f(a) = f(ba)$
Where the second equality holds because $\mathbb{Z} \times \mathbb{Z}$ is commutative. But $ab \neq ba$, hence $f$ is not injective.
Regarding 2:
Consider any $(n, m) \in \mathbb{Z} \times \mathbb{Z}$. Then note that $i_\ast (a^n b^m ) = i_\ast (a^n) i_\ast (b^m) = n i_\ast (a) m i_\ast (b) = n (1,0) + m (0,1) = (n,m)$. So $i_\ast$ is surjective.
Regarding 3:
There is a theorem as follows (Hatcher, page 36):
If a space $X$ retracts onto a subspace $A$, then the homomorphism $i_\ast : \pi_1(A, x_0)\to \pi_1(X, x_0)$ induced by the inclusion $i : A \hookrightarrow X$ is injective. If $A$ is a deformation retract of $X$ , then $i_\ast$ is an isomorphism.
In 1 you showed that $i_\ast$ is not injective hence (by contraposition) you get that $S^1 \times S^1$ does not retract onto $S^1 \vee S^1$.
Regarding 4:
There is another theorem (Hatcher page 46):
If $\varphi : X \to Y$ is a homotopy equivalence, then the induced homomorphism $\varphi_\ast :\pi_1(X,x_0)\to \pi_1 (Y,\varphi(x_0))$ is an isomorphism for all $x \in X$.
You know that the induced homomorphism isn't injective hence it cannot be an isomorphism and hence $i_\ast$ cannot be a homotopy equivalence.