I would be very appreciative if someone could check some of my understanding below, and/or answer my question below in bold.
$\def\chain{\stackrel{\partial}{\longrightarrow}}\def\chainz{\stackrel{0}{\longrightarrow}}\def\chaini{\stackrel{\sim}{\longrightarrow}}\def\cochain{\stackrel{\delta}{\longleftarrow}}$ If we consider the singular homology of a point, we have a unique $$\sigma_i:\Delta^i\to \{\text{pt}\}$$ for all $i\geq 0$. For whichever coefficients we take, we will freely generate a group with these $\sigma_i$. Say we took the coefficients $\Bbb Z$, then this chain complex: $$C_{n+1}\chain C_n \chain C_{n-1}\chain \cdots\chain C_0\chain 0$$
With our chosen generators, this gives us: $$\Bbb Z\chain \Bbb Z \chain \cdots \chain \Bbb Z \chain 0$$
Where we can see that the boundary map $\partial\sigma_n=\sum_{i}\sigma_n|_{[v_0,\cdots,\hat{v_i},\cdots,v_n]}$ and since all of these restricted chains(?) are equal to $\sigma_{n-1}$ and we have an alternating sum of $n+1$ of these, if $n$ is even, this will simply give us $\sigma_{n-1}$ and if it is odd, we will get $0$.
So we actually have: $$\cdots\chaini\Bbb Z\chainz\Bbb Z\chaini\Bbb Z\chainz\Bbb Z\longrightarrow0$$
And the homology groups are thus easily seen to be $$H_i(\{\text{pt}\})=\begin{cases}\Bbb Z,&i=0\\ 0,&i>0 \end{cases}$$
We then can dualize this taking $\delta=\partial^*$: $$\cdots\cochain C_{n+1}^*\cochain C_n^* \cochain C_{n-1}^* \cochain\cdots$$
$$\cdots\cochain \hom(\Bbb Z,\Bbb Z)\cochain \hom(\Bbb Z,\Bbb Z) \cochain \hom(\Bbb Z,\Bbb Z) \cochain\cdots$$
Where $\delta \psi=\partial^*\psi=\psi\partial:C_{n+1}\to C_n\to \Bbb Z$
So I can see indeed that we have $\delta:C_{i}^*\to C_{i+1}^*$, does a constant map dualize to a constant map? Does the zero map dualize to a constant map? In particular, how can I conclude that for the cohomology groups we have(assuming it is even true): $$H^i(\{\text{pt}\};\Bbb Z)=\begin{cases}\Bbb Z,&i=0\\ 0,&i>0 \end{cases}$$