Let X be the shrinking wedge of circles. Which is the radius of circles $X \in R^2$ such that it's the union of $C_n$ circles centered at $(\frac{1}{n},0)$ with radius $\frac{1}{n}$ for n=1,2,3...

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I'm having a bit of trouble with sheeted aspect. I'm thinking that the answer is two straight lines intersecting in a point. Does sheeted refer to covering it twice?

So like a straight line can cover that space I think.

  • $\begingroup$ `2 sheeted covering space' means that the map $f : Y \to \tilde X$ satisfies $f^{-1}(p)$ is two points for all $p \in \tilde X$. $\endgroup$ – Ryan Budney Mar 20 '12 at 23:01
  • $\begingroup$ @RyanBudney That makes sense. However, it was saying for a connected space the cardinality is constant. So wondering how does being connected preserve that property. $\endgroup$ – simplicity Mar 20 '12 at 23:10
  • $\begingroup$ Are you referring to the cardinality of $f^{-1}(p)$ being constant provided $\tilde X$ is connected? This is unrelated to your question, but yes this is true. It's a basic connectivity argument -- form a disconnection of $\tilde X$ by considering the subspaces of $\tilde X$ where $f^{-1}(p)$ have different cardinalities. $\endgroup$ – Ryan Budney Mar 20 '12 at 23:15
  • $\begingroup$ I remember this question and some answers. What have you thought of so far? You should note that two straight lines intersecting at a point is not a covering space of $\hat{X}$ at all, so that won't do. In general, as there are infinitely many earrings now, you expect to have twice as many (i.e. still infinitely many) earrings or earring representatives in your answer. If you're confused by sheets, note that $\hat{X}$ is an infinite sheeted covering of $X$, but it's still connected. $\endgroup$ – davidlowryduda Mar 20 '12 at 23:23
  • $\begingroup$ @mixedmath Well, yeah thinking you have to modify it. So you have a straight line of earrings, two straight line of earrings intersecting at a point. $\endgroup$ – simplicity Mar 20 '12 at 23:31

See this :

the composition is not a covering space because there is no open neighboorhood of the "connecting point" that pullbacks homeomorphically onto copies of itself.

For any neighboorhood you try, you will have to take a small circle in it, and this circle will be pulled apart in one of the pullbacks.

  • $\begingroup$ This is an intuition developing question! To just blurt an answer out... In addition, it's good to note that we are explicitly considering the string of Hawaiian earrings as a subspace of R2, so the topology is different than the weak CW topology. So it's not as bad as it might at first seem. $\endgroup$ – davidlowryduda Mar 20 '12 at 23:52
  • $\begingroup$ Just to check if I understand your ideas correct: we take two disjoint copies $\tilde{X}$, and at each $z\in \mathbb{Z}$ we glue the two Hawaiian ring above at $z$ by gluing the circle $C_z$, which is smaller and smaller, and the gluing map is $x\rightarrow x^2$, a double cover from $S^1$ to $S^1$. Above each $z$, the map from the glued space to $X_n$ is sending the glued circles $C_z$ to a fixed circle $C_1$. Then this is a local homomorphism since locally it is just permuting. And it has constant fibers thus a covering. $\endgroup$ – Ivan Feb 17 at 11:00
  • $\begingroup$ But the composition is not a covering for $X$, since for any small neighborhood around the connecting point, its preimage, at $z$ is small, contains two disjoint unions of the interval and when $z$ is large the circles are becoming smaller thus the whole rings are contained in the preimages. So they are not homeomorphic. $\endgroup$ – Ivan Feb 17 at 11:03

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