# Homeomorphism vs diffeomorphism in the definition of k-chain

In "Analysis and Algebra on Differentiable Manifolds", 1st Ed., by Gadea and Masqué, in Problem 3.2.4, the student is asked to prove that circles can not be boundaries of any 2-chain in $\mathbb{R}^2-\{0\}$. I understand the solution which makes use of the differential of the angle $\theta$.

The pdf at http://www.math.upenn.edu/~ryblair/Math%20600/papers/Lec17.pdf mentions that a singular k-cube $c$ is an homeomorphism of the unit k-cube.

Since discs are homeomorphic to unit squares, if $c$ is just asked to be an homeomorphism, the circle can be a boundary of a 2-chain. But a disc is not diffeomorphic to the unit square.

Is it correct to say that for the exercise to make sense, the definition of a singular k-cube to consider has to be the one using a diffeomorphism ?

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Sorry maybe I'm missing something but why do you say that "if c is just asked to be a homeomorphism, the circle can be a boundary of a 2-chain" in $\mathbb{R}^2-0$? What 2-chain do you have in mind? –  Tim kinsella Apr 7 '13 at 1:41
Not one in particular. But the fact that there are homeomorphic indicate that there exist at least one homeomorphism. And in the pdf the only requirement for a k-cube $c$ is to be an homeomorphism. $c$ could be something similar the graphical answer to this post: physicsforums.com/showthread.php?t=341907 and in that case the circle is a boundary of this 2-chain in $\mathbb{R}^2-\{0\}$. –  vkubicki Apr 7 '13 at 10:27
I just realized that the pdf requirements are even weaker than an homeomorphism as a bijection is not required. –  vkubicki Apr 7 '13 at 10:34
in that picture the square contains the origin so it's hard to see how to consider it a chain in the punctured plane. But in any case the answer to your question is no; there's nothing essentially smooth or differentiable about the fact that the unit circle is not a boundary in the punctured plane. In developing singular homology you get the same result working with chains which are formal sums of continuous maps of the standard simplex into the space under consideration. I hope I'm making sense –  Tim kinsella Apr 7 '13 at 19:53
yes exactly! from the viewpoint of algebraic topology, the removal of that one point makes the space into a circle. more precisely, the punctured plane has the "homotopy type" of a circle. –  Tim kinsella Apr 7 '13 at 23:46

As discussed in the comments, the unit circle defines a singular 1-simplex, $\sigma$ (i.e. a continuous map from the closed interval) which is not the boundary of any singular 2-chain (i.e. any formal sum of continuous maps from the standard 2-simplex) in the punctured plane. One way to see this is by noting that the punctured plane deformation retracts onto the unit circle and that $\sigma$ represents a generator for $H_1(S^1)$ (which can be seen by using a Mayer-Vietoris sequence, for instance). It should be mentioned that you can see the importance of the punctured origin in the definition of the 1-form $d\theta$ you mention, which does not extend smoothly to the entire plane.