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): $$H^i(\{\text{pt}\};\Bbb Z)=\begin{cases}\Bbb Z,&i=0\\ 0,&i>0 \end{cases}$$

  • $\begingroup$ Yes, constants go to constants. $\endgroup$ – Pedro Tamaroff Oct 18 '16 at 11:10

Your singular complex for a point is correct, it is $$\tag{*} \cdots \to \mathbb{Z} \xrightarrow{1} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{1} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0$$

Cohomology (with coefficients in $A$ in general) is by definition obtained by applying the contravariant functor $\operatorname{Hom}_\mathbb{Z} (-, A)$ to the singular complex $C_\bullet (X)$. It dualizes the differentials $d_n\colon C_n\to C_{n-1}$ to maps going in the other direction $d^{n-1}\colon C_{n-1}^\vee \to C_n^\vee$ (which we renumber: to $d_n$ corresponds $d^{n-1} = d_n^\vee$). The cohomology of the resulting (cohomological) complex is by definition $H^\bullet (X,A)$.

To answer your particular question, it should be clear from the definitions that after taking $\operatorname{Hom}_\mathbb{Z} (-,A)$, the multiplication by $n$ map $\mathbb{Z}\xrightarrow{\times n} \mathbb{Z}$ corresponds again to the multiplication by $n$ map $A\xrightarrow{\times n} A$.

In particular, using this, you can write down the complex $\operatorname{Hom}_\mathbb{Z} (C_\bullet (pt), \mathbb{Z})$, which will be essentially the same thing as (*), only with the arrows reversed.

To give a more general picture, note that since $H^\bullet (X,A)$ is obtained by some formal "dualization" of the complex $C_\bullet (X)$, the relation between $H_\bullet (X)$ and $H^\bullet (X,A)$ boils down to some problem in homological algebra: what happens to the homology of a complex when we apply $\operatorname{Hom}_\mathbb{Z} (-, A)$ to it. The answer is the universal coefficient theorem, which says that you have a short exact sequence $$0 \to \operatorname{Ext}^1_\mathbb{Z} (H_{n-1} (X), A) \to H^n (X,A) \to \operatorname{Hom}_\mathbb{Z} (H_n(X), A) \to 0$$ which is split non-canonically, i.e. you have a non-canonical isomorphism $$H^n (X,A) \cong \operatorname{Hom}_\mathbb{Z} (H_n(X), A) \oplus \operatorname{Ext}^1_\mathbb{Z} (H_{n-1} (X), A).$$

In particular, if all $H_\bullet (X)$ are free abelian groups, the $\operatorname{Ext}^1$ will be zero, so we obtain isomorphisms $$H^n (X,A) \cong \operatorname{Hom}_\mathbb{Z} (H_n(X), A)$$ so that cohomology is $\mathbb{Z}$-dual of homology.

This agrees with your particular example.


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