7
$\begingroup$

This is problem 1.2.20 from Hatcher's algebraic topology.

enter image description here

I am able to complete the first part. But I have trouble proving they are homotopy equivalent but not homeomorphic. Can someone please give some hints? Thanks.

$\endgroup$
11
  • $\begingroup$ Maybe you will find interesting math.stackexchange.com/questions/69698/… $\endgroup$
    – mfl
    Jul 10, 2017 at 14:11
  • $\begingroup$ I just found out that my proof for part one is wrong... $\endgroup$
    – JSCB
    Jul 10, 2017 at 14:53
  • $\begingroup$ Let me call the wedge $Y$. To define $g:Y\to X$, fix a homeomorphism $h$ from $S^1$ to the line $x=1$ in the Riemann sphere. Send the $n$-th copy of $S^1$ in the wedge to the line $x=\frac{1}{2n}$ in the Riemann sphere, by using $h$ and then mapping the line $x=1$ to the line $x=\frac{1}{2n}$ by translation. Finally send all the lines $x=\frac{1}{2n}$ to X by doing inversion in the Riemann sphere with respect to the unit circle. To define $f:X\to Y$ send the point $nz+n$ with $n\in\mathbb{N}$ and $|z|=1$ to the point $z$ in the $n$-th copy of $S^1$ in the wedge. $\endgroup$
    – Bettybel
    Jul 10, 2017 at 18:24
  • 2
    $\begingroup$ @GaryD $\bar{X}=X\cup (\{0\}\times \Bbb R)$. The map $\bar{X}\to X$ that fixes $X$ and sending the $y$-axis to the origin is not continuous. $\endgroup$
    – user302934
    Oct 22, 2020 at 7:16
  • 5
    $\begingroup$ Does this answer your question? The union of growing circles is not homeomorphic to wedge sum of circles $\endgroup$ Oct 24, 2020 at 3:24

1 Answer 1

9
+50
$\begingroup$

Comparing the topology of the nested-circle space X and the wedge-sum space Y.

Not homeomorphic

Here's the visual intuition. The problem for homeomorphism is that the wedge sum makes all the circle-copies $S^1$ independent of one another. You can make an open set in $Y$ by choosing one arc (such as the red curve, above) independently from each of the circles; the union of arcs is an open set.

In contrast, the circles of $X$ are clustered together at the origin. Any open neighborhood of the origin will intersect all of the circles, and the resulting arc lengths (blue curves) are not independent of each other. Instead, we know that the neighborhood contains an open ball around the origin, and so each curve must be at least as long as the ball's radius.

This is the fundamental difference between the two topologies: In both $X$ and $Y$, the neighborhoods of the origin include families of arcs, one per circle. In $Y$, the arcs from different circles may have arbitrary lengths. In $X$, the arcs always have some lower limit $r$.

You can use this intuition to prove that $X$ is first countable (it's a subspace of a metric space $\mathbb{R}^2$) but $Y$ is not; hence they're not homeomorphic. ($Y$ is not first countable: for any collection $B_1, B_2, \ldots,$ of neighborhoods of the origin, you can find for each $B_k$ a "red arc" $A_k$ in the $k$th circle that is sufficiently small that it does not contain $B_k$; from the combination of these arcs, you can obtain a neighborhood $ \bigvee A_k$ of the origin that (by diagonal argument) contains none of the $B_k$; hence the collection $B_1,B_2,\ldots$ isn't a neighborhood basis. It can be useful to think about why this construction doesn't transfer over to $X$.)

...but homotopy equivalent

The intuition that $X$ and $Y$ are homotopy equivalent: note that you can construct either space following the same construction technique: start with an infinite disjoint union of circles $\coprod_\infty S^1$, select one point from each circle, and glue the selected points together. In other words, $X$ and $Y$ are cell complexes made of the same parts and constructed in the same way.

Here's one way to construct a homotopy equivalence explicitly. There's a bijective map $f:Y\rightarrow X$ that sends circles to circles in the obvious way, mapping the origin onto the origin. Roughly speaking, it is continuous because $Y$ includes all the same open sets as $X$ (plus some more).

Next, in the space $X$, consider a small closed ball around the origin, say of radius $r=1/4$. There is a continuous map $\alpha:X\rightarrow X$ which shrinks all points in the ball to the origin, pulling the rest of each circle closed. This map is continuous and moreover homotopic to the identity (you can gradually shrink the ball to a point.)

A bouquet of circles, with a circular region quotiented to a point.

There is a bijective function $\tilde{g}:X\rightarrow Y$ which sends circles to circles, mapping the origin onto the origin. It is not continuous, because there's no analogue in $X$ to the open neighborhoods made of arbitrary-sized arcs through the origin in $Y$. In contrast, the composition $g \equiv \tilde g\circ \alpha$ repairs this problem— it is continuous. Indeed, then the inverse image of every neighborhood of the origin in $Y$ contains the previously shrunk-to-zero ball of radius 1/4.

The pair $f$ and $g$ comprise the homotopic equivalence. The composition $f\circ g:X\rightarrow X$ is just the map $\alpha$ again, which shrinks the $1/4$-radius ball to zero. The composition $g\circ f:Y\rightarrow Y$ performs an analogue of that quotienting operation on $\bigvee_\infty S^1$, sending various arcs of the origin, one from each circle, to the base point. It is similarly homotopic to the identity.

$\endgroup$
4
  • $\begingroup$ The map $\bar{X}\to X$, sending the $y$-axis to the origin, and fixing $X$ is not continuous, I think, because $\bar{X}$ is the closure of $X$. $\endgroup$
    – user302934
    Oct 22, 2020 at 7:15
  • $\begingroup$ The set $Y$ is not open in $\bigvee_1^\infty S^1$, because in particular it is not open when intersected with each of its subspaces $S^1$. You would still need a neighbourhood in each copy of $S^1$. The difference between $X$ and $Y$ is that those neighbourhoods in $Y$ can become arbitrarily small. $\endgroup$
    – abhi01nat
    Oct 22, 2020 at 8:51
  • $\begingroup$ @abhi01nat Thank you; you're right. The preimage of the red arc under the quotient map from $\coprod_\infty S^1$ is not an open set. How to draw a better picture? $\endgroup$
    – user326210
    Oct 22, 2020 at 9:01
  • $\begingroup$ I'm not sure; one way perhaps is to note that any neighbourhood of the basepoint in $X$ will contain a "minimum" ball, that is each circle has an open set of at least a given size. This is not true in $Y$ where the neighbourhoods can be of arbitrary sizes, so the picture can show smaller and smaller arcs shrinking to a point. $\endgroup$
    – abhi01nat
    Oct 22, 2020 at 9:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.