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 $1$-chains 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?
Best Answer
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 $1$.