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Trying to solve this problem $1.3.7$ from Hatcher, but the description already confuses me. I know if we didn't include the segment connecting $y = \sin(\frac{1}{x})$ and segment $ [-1,1]$ in $y$ direction we have $2$ path components - basically segment is one component, and we know topologist's sine curve is not path connected. Also see here.

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I am assuming what does quotient map here mean is that after collapsing the segment in $y$ direction, we get a circle. But, if you remove a single point on the arc connecting, you get $2$path components , since contracting the remaining part of arc we get topologist's sine curve. On the other hand, removing a point in circle leads to connected space.

What am I doing wrong here?

Rias Gremory
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  • I don't really understand your question. Firstly, the topologists sine curve has two path components, but only one component. Regardless, what does that have to do with removing a point from the circle? The circle is the quotient of $Y$, it is not homeomorphic to $Y$. – M W Apr 03 '24 at 21:54
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    @MW Rephrasing: "Let $Y'$ be the quotient space. Take out a point from the image of the arc in $Y$. This space deformation retracts onto the topologist's sine curve which has 2 path components. If $Y'$ where homeomorphic to $S^1$, then taking out a point should decompose $S^1$ into 2 path components as well, but this is not the case." Apart from changing components into path components I don't think the question is unclear. – Ben Steffan Apr 03 '24 at 22:04
  • I wrote component even though I meant path component , fixed now. What I mean is suppose you collapsed the segment. Then removing a point, we have $2$ path components, but if collapsed $Y$ was circle, removing a point shouldn't have $2$ path components – Rias Gremory Apr 03 '24 at 22:05
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    @MahammadYusifov "we get the topologist's sine curve" are you sure about this? :) You might want to think about the quotient topology here for a second. – Ben Steffan Apr 03 '24 at 22:07
  • @BenSteffan if you let $Y'$ be the quotient space then it doesn't deformation retract onto a topologists sine curve anymore. – M W Apr 03 '24 at 22:07
  • @MW Exactly, whence my last comment :) – Ben Steffan Apr 03 '24 at 22:08
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    @BenSteffan gotcha, re-reading I guess its clear he's talking about the quotient - I originally just assumed he meant the original space because of reason we both gave. – M W Apr 03 '24 at 22:12
  • The mistake is in the sentence "But, if you remove a single point on the arc connecting, you get 2 path components, since contracting the remaining part of arc we get topologist's sine curve." The sentence is poorly structured. Try to break it into several carefully written parts and your mistake will become clear. – Moishe Kohan Apr 04 '24 at 01:00

1 Answers1

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Quotients are not homeomorphisms. It is true that if we take the quasicircle $Q$ and delete a point on the connecting arc $-1\to1$, the resulting space $Q'$ has two path components (and one component). If $q:Q\to S^1$ is the quotient map under consideration, it is also true that $q(Q')$ has one path component. ... That's not a problem!

You cannot prove that quotients identify path components $1-1$; it is perfectly possible for a quotient to glue two path components into one. For a very basic example consider $[-2,-1]\cup[1,2]$ quotient by $\{-1,1\}$; there are two path components, but the quotient is a line and is path connected. Here what's going on is that the quotient map brings the two components of $Q'$ together in collapsing the sine curve down to a point, if you want to visualise it. It removes the 'weirdness' at the endpoint of the sine curve by collapsing everything down.

There is no problem!

FShrike
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