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I am considering homology and cohomology with integer coefficients.

For singular homology of a topological space $X$, we know that $H_0(X)$ is free on the number of path-components of $X$, and $\widetilde H_0(X)$ (corresponding to the homology of the augmented chain complex $\dots \to C_1(X) \to C_0(X) \to \mathbb{Z} \to 0$, where the map $\epsilon : C_0(X) \to \mathbb{Z}$ sends generators $\sigma$ to $1\in \mathbb{Z}$) satisfies the formula $H_0(X) \cong \widetilde{H}_0(X) \oplus \mathbb{Z}$.

My questions are regarding singular cohomology. First, is the group $H^0(X)$ always (or ever) a free abelian group? Maybe if $X$ has finitely many components it is free? What about $\widetilde{H^0}(X)$? (Where $\widetilde{H^0}(X)$ is the degree 0 cohomology of the dual of the augmented complex $\dots \to C_1(X) \to C_0(X) \to \mathbb{Z} \to 0$) Is $\widetilde{H^0}(X)$ ever (or always?) free?

My second question is whether the formula $H^0(X) \cong \widetilde {H^0}(X) \oplus \mathbb{Z}$ holds in general, or at least possibly for spaces $X$ with finitely many path components. I can show using a long exact sequence of homology that there is an exact sequence $0 \to \mathbb{Z} \to H^0(X) \to \widetilde{H^0}(X) \to 0$, but I am not sure if this sequence splits or not (since I don't know if $\widetilde{H^0}(X)$ is free).

Thanks for input on this, I think it is probably a dumb question, but I can't seem to figure it out. I think $\widetilde{H^0}(X)$ is "functions which are constant on path components, modulo constant functions", so intuitively it seems like $\widetilde{H^0}(X)$ could be free. On the other hand, I think $C^0(X) = \hom(C_0(X),\mathbb{Z}) \cong \prod \hom(\mathbb{Z},\mathbb{Z}) \cong \prod \mathbb{Z}$ may not usually be free, so it seems like $H^0(X)$ might not be free.

user620793
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Here is the situation. You have a set $P$ of path-components, and know that $H^0(X) = \Bbb Z^P$, the set of functions on $P$. This is only free when $P$ is finite.

This contains $\Bbb Z$ as the constant functions. You are looking to construct a section $s: \Bbb Z^P/1 \to \Bbb Z^P$. By picking an arbitrary element of $\Bbb Z^P$ and sending it to $x - s(x)$, you obtain a retraction $r: \Bbb Z^P \to \Bbb Z$. In fact, these are equivalent: given such a retraction the appropriate section is given by taking an arbitrary lift $x$ of $y$ and sending $y$ to $y - r(y)$.

Inspired by this, you pick a random point $p \in P$ and set $r(f) = f(p)$, just evaluation at that point. This of course satisfies the property that $r(1) = 1$, as desired. And so we see that your map does have a section, defined by sending $[f] \in \Bbb Z^P/1$ to the unique function on $P$ which differs from $f$ by a constant and has $f(p) = 0$. If $P' = P \setminus p$, we have identified $$\tilde H^0 = \Bbb Z^P/1 = \Bbb Z^{P'}.$$

Note that we did not need $\Bbb Z^P/1$ to be free. It won't be unless $P$ is finite.