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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.

  1. 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.

  2. 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 $i_*$?

  3. 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?

  4. 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!

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For 1) and 2), it might help to write down some elements of $\pi_1(S^1 \wedge S^1)$ and then calculate their images under $i_*$. Is this homework? –  Adam Saltz Dec 26 '11 at 14:54
    
Thank you for your help! No, I am on Christmas break. –  user21787 Dec 26 '11 at 15:08

2 Answers 2

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.

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Dear Matt: I counted $25$ users with the username Matt or matt. I was wondering if you couldn't do something, like adding an initial, to make your username unique. - This would be just to help your readers... –  Pierre-Yves Gaillard Dec 26 '11 at 15:46
    
Thank you for taking the time to answer my question this has been very helpful for my understanding the problem! –  user21787 Dec 26 '11 at 15:46
    
@user21787 My pleasure! : ) –  Rudy the Reindeer Dec 26 '11 at 15:49
    
Dear @Pierre-YvesGaillard: I will see to it. –  Rudy the Reindeer Dec 26 '11 at 18:45
    
@Pierre-YvesGaillard: Better? : ) –  Rudy the Reindeer Jan 21 '12 at 13:54

pick a point on the torus, and consider the two circles through the point, one going "vertically" and the other "horizontally" (draw a picture). the inclusion map is not an isomorphism but it is surjective. the kernel is generated by the commutator of the two loops. try drawing the standard square-with-sides-identified (the top and bottom one of the generating loops, the left and right sides the other, all four corners our base point) to see that the commutator is trivial (the "square" is basically the homotopy from $aba^{-1}b^{-1}$ to the constant map). the spaces are not homotopic as they have different invariants, namely $\pi_1$ as you noted.

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Thanks a lot I appreciate the help! –  user21787 Dec 26 '11 at 15:47

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