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Continuum theory, point-set topology, spaces with algebraic structure, foundations, dimension theory, local and global properties.
4
votes
Accepted
Are open sets determined by paths?
A space $X$ is called $\Delta$-generated if $U$ is open in $X$ if and only if $\alpha^{-1}(U)$ is open in $[0,1]$ for every path $\alpha:[0,1]\to X$.
It's easy to see that a space $X$ is $\Delta$-gen …
7
votes
Which spaces have enough curves
A space $X$ whose topology agrees with the final topology with respect to all maps $I\to X$ is often called a delta-generated ($\Delta$-generated) space. The category of $\Delta$-generated spaces is a …
10
votes
Accepted
Fundamental groups and homology groups of closed subsets of the plane
Fundamental Group: The fundamental group of a planar set naturally injects into the first Cech homotopy group, which is an inverse limit of free groups. In particular, the algebraic restrictions gaine …
2
votes
covering theory with compact open topology
$p^{\ast}$ may not be surjective without extra assumptions. For example, you can check out Zeeman's example in Spanier's textbook of non-equivalent coverings corresponding to the same subgroup of $\pi …
3
votes
Accepted
Let $U$ be a simply connected open subset of ${\Bbb S}^2$, is the complement of $U$ also sim...
$F=S^2\backslash U$ need not be path connected, e.g. $F$ could be homeomorphic to the closed topologists sine curve. However, every path component of $F$ must be simply connected.
By identifying $U$ …
37
votes
Is there a "universal" connected compact metric space?
There is no such continuum. See
Z. Waraszkiewicz, Sur un problème de M.H. Hahn, Fund. Math. 22 (1934) 180–205.
Waraszkiewicz constructed an uncountable family $W$ of continua in the plane called Wa …
15
votes
Accepted
when is a locally homeo a covering map?
This answer takes a more general viewpoint than Alexandre's. The generality is in response to the small number of assumptions on the spaces involved.
First, you should assume that $Y$ is locally path …
7
votes
Accepted
Lifts across covering maps
Suppose you have basepoints $x_0\in X$, $z_0\in Z$ and $p(z_0)=f(x_0)$. The lift $\tilde{f}:X\to Z$ such that $p\circ \tilde{f}=f$ exists and is continuous if and only if
1) $f_{\ast}(\pi_1(X,x_0))\ …
1
vote
Looking for general approaches to show connectedness of topological groups
Perhaps this is helpful for the locally compact case.
Corollary 3.1.12 in "Topological groups and related structures" by Arhangel'skii and Tkachenko gives a nice characterization of connectedness in …
4
votes
Is the wedge sum of two cones over the hawaiian earring contractible?
The space you are talking about is sometimes called the Griffiths space. As Henry Horton suggests, to prove it is not contractible you can show the fundamental group is non-trivial (even though its Ce …
8
votes
Accepted
Construction of the universal covering space via compact-open topology
Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path …
4
votes
Accepted
fundamental groups of complements to countable subsets of the plane
Thanks to the comments, my original cardinality bound $\aleph_1\leq |S|\leq \mathfrak{c}$ has been refined to the equality $|S|=\mathfrak{c}$ that I originally suspected.
For Question 1: $S$ has the …
6
votes
Accepted
Homotopy equivalent fibers and Fibrations
The figure below gives a simple but extreme counterexample, which I think has all the lifting properties one might want except for actually being a true fibration. The map is the identity everywhere e …
22
votes
Accepted
Universal covering space for non-semilocally simply connected spaces
My answer to this question gives an example of a locally path connected (but non-semilocally simply connected) space $HA\subset\mathbb{R}^3$ called the Harmonic archipelago: draw the Hawaiian earring …
14
votes
Accepted
In a subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple loop t...
One-dimensional metric spaces and planar sets do have the property that you're interested in. To explain why this works out in such generality requires a combination of planar topology, continuum theo …