How to represent a generator of $H^1(S^1,\mathbb{Z})$ by a singular cocycle? It is well-know that $H^1(S^1,\mathbb{Z})\cong \mathbb{Z}$. It is easy to represent such an element in de Rham cohomology but I wonder how to represent the element $1\in \mathbb{Z}\cong H^1(S^1,\mathbb{Z})$ as a singular cocycle.
Let $C_1(S^1)$ be the free abelian group generated by singular $1$-cochains of $S^1$, i.e. generated by continuous maps $\mu: \Delta^1\to S^1$. A singular $1$-cochain is a homomorphism
$$
\phi: C_1(S^1)\to \mathbb{Z}.
$$
$\phi$ is a cocycle means that for a singular $2$-chain $\sigma: \Delta^2\to S^1$ we have
$$
\phi(d_0\sigma)-\phi(d_1\sigma)+\phi(d_2\sigma)=0.
$$
Moreover $\phi$ represents the element $1$ means
$$
\phi(S^1)=1
$$
where $S^1$ denote the singular $1$-chain of one full circle.
I found that it is difficult to choose values of $\phi$ on all singular $1$-chains because we don't allow non-integer values. Do we need to use the axiom of choice here?
 A: The method that you gave in your own answer only applies to smooth singular simplices $\sigma$, because the pullback operator $\sigma^*$ is not defined when $\sigma$ is an arbitrary singular simplex which is merely continuous.
But your method can be adapted to arbitrary singular simplices like this.
Start by taking any function $f : \mathbb R \to \mathbb Z$ having the property that $f(x+n)=f(x)+n$ for all $x \in \mathbb R$; for instance, you can take $f(x) = \lfloor x \rfloor$, the greatest integer function. (By the way, no axiom of choice is needed for writing down this one function $f$).
Consider the universal covering map $p(x)=\exp(2\pi i x)$. For any singular $1$-simplex $\sigma : [0,1] \to S^1$, covering space theory gives you, for each $t_0 \in \mathbb R$ such that $p(t_0)=\sigma(0)$, a unique continuous $\tilde\sigma : [0,1] \to \mathbb R$ such that $\tilde\sigma(0)=t_0$ and $p(\tilde\sigma(t))=\sigma(t)$ for all $t \in [0,1]$. Define
$$\phi(\sigma) = f(\tilde\sigma(1))-f(\tilde\sigma(0)) \in \mathbb Z
$$
This is well-defined independent of the choice of $t_0$, because for any other choice $t'_0$ determining an alternative lift $\tilde\sigma' : [0,1] \to \mathbb R$, one has $\tilde\sigma'(t)-\tilde\sigma(t) \in \mathbb Z$. But a continuous function with values in $\mathbb Z$ takes a constant value $n$. It follows that $f(\tilde\sigma'(0))=f(\tilde\sigma(0))+n$ and $f(\tilde\sigma'(1))=f(\tilde\sigma(1))+n$, and so
$$f(\tilde\sigma'(1)) - f(\tilde\sigma'(0)) = f(\tilde\sigma(1)) - f(\tilde\sigma(0))
$$
One can prove that this cochain is indeed a cocycle, by taking any singular 2-simplex $\sigma : \Delta^2 \to S^1$ and again use covering theory to lift it to a continuous map $\tilde\sigma : \Delta^2 \to \mathbb R$ where the lift is determined by any single one of its values. The key idea is that one of the lifts of the $i^{\text{th}}$ face map $\sigma \mid \partial_i\Delta^2$ is $\tilde\sigma \mid \partial_i\Delta^2$, and from this one easily calculates that the cocycle equation holds:
\begin{align*}
\phi(\sigma \mid \partial_0 \Delta^2) - \phi(\sigma &\mid\partial_1\Delta^2) + \phi(\sigma\mid\partial_2\Delta^2) \\ &= \bigl(f(\tilde\sigma(v_2)) - f(\tilde\sigma(v_1))\bigr) \\
& \quad- \bigl(f(\tilde\sigma(v_2)) - f(\tilde\sigma(v_0))\bigr) \\
& \quad\quad + \bigl(f(\tilde\sigma(v_1)) - f(\tilde\sigma(v_0))\bigr) \\ &= 0
\end{align*}
And, finally, $\phi$ applied to $\sigma(t) = (\cos(2\pi t),\sin(2\pi t))$ clearly outputs $1$.
A: I found a construction with the help of differential forms: First identify $S^1$ with the interval $[0,2\pi)$ (not a homeomorphism). Let $d\theta$ be the usual differential form on $S^1$. For any $1$-chain $\sigma: \Delta^1\to S^1$, we have
$\sigma(0)\in [0,2\pi)$ and $\sigma(1)\in [0,2\pi)$, where $0$ and $1$ are the two end points of $\Delta^1$.
Then we define a $1$-cochain $\phi: C_1(S^1)\to \mathbb{Z}$: for any  $1$-chain $\sigma: \Delta^1\to S^1$, we have
$$
\phi(\sigma)=\frac{1}{2\pi}\Big(\int_{\Delta^1}\sigma^*(d\theta)-\sigma(0)+\sigma(1)\Big).
$$
It is clear that $\phi(\sigma)\in \mathbb{Z}$ and $\phi$ is a $1$-cocycle. Moreover
$$
\phi(S^1)=1.
$$
A: You will find in Lang's book on complex analysis a definition of the integral $\int_\gamma f(z)\mathrm d z$ of a holomorphic function $f$ defined on an open subset $\Omega$ of the complex plane along a continuous curve $\gamma:[0,1]\to\Omega$.
In particular, this gives you a morphism of groups $\nu:C_1(S^1)\to\mathbb C$ that on any continuous $\gamma:[0,1]\to S^1$ takes the value $\nu(\gamma)=\frac{1}{2\pi i}\int_\gamma\mathrm dz/z$. Information about integrals that you can find in that book implies that this map vanishes on $1$-coboudaries (Lang proves Cauchy's theorem in a form very close to this, but using winding numbers, so one has to work a little), and that it is integer valued on $1$-cycles. This is the map you want.
The difficulty of this, of course, is proving that one can integrate holomortphic functions along continuous curves.
