I came a cross an exercise the other day considering the following quotient space: Let $T$ be a torus and let $A, B \hookrightarrow T$ be two parallel circles. Let $X$ be the quotient space collapsing all of $A$ to a point, and all of $B$ to a different point.

What would be the geometric interpretation of $X$? I was thinking something along the lines of a torus pinched at two points? Any ideas?


2 Answers 2


Yes, it would be a "double-pinched" Torus, and here is a picture even though I know images are discouraged.

Also, this was discussed in some detail in this post:

Homology of doubly-pinched torus

enter image description here

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    $\begingroup$ Images are discouraged when an OP uses them in lieu of actually typing the question. They are entirely appropriate in both questions and answers when they contribute to the understanding. $\endgroup$ Dec 4, 2023 at 15:56
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    $\begingroup$ Not only they are appropriate. In some cases one image can replace thousands words. This is the case. The answer could literally be the image only, and it would still be perfect. $\endgroup$
    – freakish
    Dec 4, 2023 at 15:59
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    $\begingroup$ Thanks (both comments) - I will not "apologize" next time - images are definitely invaluable in these cases! $\endgroup$
    – AlgTop1854
    Dec 4, 2023 at 16:00
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    $\begingroup$ I think it's just screenshots of text that are discouraged, not images in general! $\endgroup$
    – Stef
    Dec 4, 2023 at 16:56

$X$ is homotopy equivalent to $S^1\vee S^2\vee S^2:$ enter image description here

$$H_0(X)=\Bbb Z, \pi_0(X)=0$$ $$H_1(X)=\Bbb Z, \pi_1(X)=\Bbb Z$$ $$H_2(X)=\Bbb Z\oplus\Bbb Z, \pi_2(X)=\Bbb Z\oplus\Bbb Z$$ For $k\geq3$: $$H_k(X)=0$$ $$\pi_k(X)=\pi_k(S^2)\oplus\pi_k(S^2)\oplus ...$$

  • $\begingroup$ @AlgTop1854 Thanks $\endgroup$
    – Bob Dobbs
    Dec 4, 2023 at 17:53
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    $\begingroup$ @BobDobbs I’m looking again - aren’t the homotopy groups of a wedge far more complicated? $\endgroup$
    – AlgTop1854
    Dec 6, 2023 at 2:31
  • $\begingroup$ @AlgTop1854 Ooops. İt seems so. math.stackexchange.com/questions/913022/… $\endgroup$
    – Bob Dobbs
    Dec 6, 2023 at 18:02
  • $\begingroup$ @AlgTop1854 İs $\pi_2$ okay? $\endgroup$
    – Bob Dobbs
    Dec 6, 2023 at 18:22
  • $\begingroup$ Yes - for $pi_2$ it matches $H_2$ by the Hurewicz theorem, and for higher homotopy groups there are additional summands (lots of them!). $\endgroup$
    – AlgTop1854
    Dec 6, 2023 at 18:38

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