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Math Help - Homology groups, homotopy equivalence and one point compactification

  1. #1
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    Homology groups, homotopy equivalence and one point compactification

    Let  X=  \textbf{R}^2 - \{ a,b \} and Y = S^1 \times  S^1 - \{c \} .

    Do the spaces X and Y have the same homology groups? Are they homotopy equivalent? Are they Homeomorphic? Do they have homeomorphic one-point compactifications?

    First, I was wondering if the space Y is homeomorphic to a cylinder? To me it looks like it's a Torus where one circle S^1 has been cut off. Or maybe, it is a Torus, where only point is missing?

    I know that H'_i (S^n) =  \textbf{Z} if  i=n and 0 otherwise. (Where H'_i (X) is the i-dimensional reduced homology of the space X)
    Also I know as an application of the Mayer-Vietoris sequence, that If S is a subspace of S^n homeomorphic to S^k for  0 \leq k < n , then H'_i ( S^n -S) =  \textbf{Z} if i=n-k-1 and 0 otherwise.

    Also, I know that  H'_i (X) = H_i (X) for i>0 (Where H_i(X) is the i-dimensional homology group of X) and that H_0(X) = H'_0(X) \oplus \textbf{Z}

    But does any of this help answer the questions? Or is it totally irrelevant in this case? If this doesn't help, how do you go about answering the questions?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by math8 View Post
    Let  X=  \textbf{R}^2 - \{ a,b \} and Y = S^1 \times  S^1 - \{c \} .

    Do the spaces X and Y have the same homology groups? Are they homotopy equivalent? Are they Homeomorphic? Do they have homeomorphic one-point compactifications?

    First, I was wondering if the space Y is homeomorphic to a cylinder? To me it looks like it's a Torus where one circle S^1 has been cut off. Or maybe, it is a Torus, where only point is missing?

    I know that H'_i (S^n) =  \textbf{Z} if  i=n and 0 otherwise. (Where H'_i (X) is the i-dimensional reduced homology of the space X)
    Also I know as an application of the Mayer-Vietoris sequence, that If S is a subspace of S^n homeomorphic to S^k for  0 \leq k < n , then H'_i ( S^n -S) =  \textbf{Z} if i=n-k-1 and 0 otherwise.

    Also, I know that  H'_i (X) = H_i (X) for i>0 (Where H_i(X) is the i-dimensional homology group of X) and that H_0(X) = H'_0(X) \oplus \textbf{Z}

    But does any of this help answer the questions? Or is it totally irrelevant in this case? If this doesn't help, how do you go about answering the questions?
    So you're ok with the alg. top. part? Assuming you mean Y=\mathbb{S}^1\times\left(\mathbb{S}^1-\{c\}\right) then X\not\approx Y. This is evident since X isn't compact and Y is. Also, I'm fairly sure (though don't quote me, I haven't checked it) that X_{\infty}\not\approx Y_{\infty} because note that since Y is already compact that \{\infty\} is open (since \{\infty\}=Y_{\infty}-\overbrace{Y}^{\text{compact}} and so Y_{\infty} is not connected. That said, I'm fairly sure X_{\infty} is.
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    Senior Member Tinyboss's Avatar
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    Y is homeomorphic to a cylinder if it's supposed to be S1 x (S1 - {c}), but I don't think that's how I'd read it, but rather as the torus minus one point. In the latter case, both spaces are homotopy equivalent to...what? Implying what about their homology groups?

    Drexel28, I don't think Y is compact under the interpretation you mentioned.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Tinyboss View Post
    Y is homeomorphic to a cylinder if it's supposed to be S1 x (S1 - {c}), but I don't think that's how I'd read it, but rather as the torus minus one point. In the latter case, both spaces are homotopy equivalent to...what? Implying what about their homology groups?

    Drexel28, I don't think Y is compact under the interpretation you mentioned.
    Oops. That was careless haha.

    Am I crazy, or is X_{\infty}\approx Y?
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  5. #5
    Senior Member Tinyboss's Avatar
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    I don't think so, since Y is not compact under either of the possible interpretations. I suspect it's meant to be the punctured torus and the twice-punctured plane, which are homotopy equivalent but not homeomorphic (since the one point compactification of the punctured torus is obviously the torus, but the 1pc of the twice-punctured plane is not a torus). Off the top of my head, I think the 1pc of the twice-punctured plane is probably homotopic to a wedge of three circles, but that's just a guess.
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    Thanks for your answer, and yes, you're right, Y is supposed to be the punctured Torus in this case.

    I would say if X is homotopy equivalent to Y then, X and Y have isomorphic homology groups. But, how do we see they're homotopy equivalent? I read somewhere that the punctured Torus is Homotopy equivalent to the bouquet of two circles, why? Can we say that the twice punctured plane is also homotopy equivalent to the bouquet of two circles?

    Also, If I understood well the reason why X and Y are not homeomorphic, is it because they have different one point compactifications? would you conclude in that case that their 1pc are non homeomorphic?
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    Let the rectangle D=[-1,1]^2 be the fundamental polygon of the torus T^2, and let \pi be the quotient map that identifies opposite edges, \pi(D)=T^2.
    The bouquet of two circles, E, is then homeomorphic to \pi(\partial{D}), the image of the boundary of D.

    Choose 0=(0, 0) to be the removed point. Along the radial beams you can define a deformation retract as follows. Given a point p \in D other than 0, Let r(p) be the intersection of the beam 0p with \partial{D}. Define F: [0,1] \times T^2 \to T^2.  F(t,p)=(1-t)*p+t*r(p). It's easy to verify that F is a strong deformation retract. This proves that the puctured torus is homotopy equivalent to the bouquet of two circles.

    Since the plane is homemorphic to an open disk, you can define similarly a deformation retract, to show that a twice punctured open disk is homotopy equivalent to the bouquet of two circles.
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    A plane with 2 points deleted is homeomorphic to a sphere with 3 points deleted. So any small open neighborhood of \infty of X_{\infty} has 3 sheets, attached together by \infty. While Y_{\infty} is a torus, where there is an open neighborhood that is homeomorphic to R^2 for each point. This shows why X_{\infty} is not homeomorphic to Y_{\infty}. Neither for X and Y.

    While X_{\infty} and Y_{\infty} have the same fundamental group Z \times Z. To see it, draw a small triangle ABC on a sphere, it is easy to construct a homotopy so that the path ABC is equivalent to the path AC. Identify the 3 points A, B and C we get X_{\infty}, with AB, BC and CA turning into circles. But there are only 2 generators( corresponding to the image of AB and BC) for the fundamental group of X_{\infty}, but not 3, as Tinyboss guessed above.
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  9. #9
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    Thank you, that helps a lot!
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