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 ?

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.
