Understanding relative cohomology of the interval I want to understand relative cohomology. The easiest example I could think of was $H^*(I,\partial I;G)$ where $I=[0,1]$ and $G$ is an abelian group.
I know the following facts: $H^*(I,\partial I)$ consists of cochains which vanish on $\partial I$, but I am not sure, what exactly this means (see later). I also know that the quotient map $q$ induces an isomorphism of cohomology groups $H^*(I,\partial I)=\tilde H^*(I/\partial I)=H^*(I/\partial I) =H^*(S^1)$. And there is a short exact sequence:
$$0\to H^0(I)\to H^0(\partial I) \stackrel \delta\to H^1( I,\partial I)\to 0.$$
Hence $\delta$ induces an isomorphism $\tilde\delta:H^0(\partial I)/H^0(I)\to H^1(I,\partial I)$.
I think a more direct proof without using the exact sequence would give me a better understanding.
For this I need an explicit description of $H^1(I,\partial I)$.
I already know that $\delta$ does. It takes a locally constant function $f:H_0(\partial I)\to G$ and maps it to $\delta f= f\partial$.
So if $f(0)=s$ and $f(1)=t$, then $\delta f(I)=f(\partial I)=f(1)-f(0)=t-s$.  
It is clear that this map has kernel $H^0(I)$, the constant functions. 
But why is it surjective? Why is it even well defined? Isn't every 1-cocycle on $I$ already a coboundary? As I said before, I need an explicit description of $H^1(I,\partial I)$. So please give me the generator $g$ of this group. Also, why is the generator $g$ the pull-back $g=q^*s$ of the generator $s$ of $H^1(S^1)$ under the quotient map $q:I\to S^1$, and where does orientation enter.
 A: I find cohomology easier to think about as a dual of homology, so I will explicitly describe that case first (specifically I will be using singular homology). I'll also just talk about the $G=\mathbb{Z}$ case for simplicity. I will still use the long exact sequence of the pair to do computations because that's sort of the whole idea of algebraic topology: not using these tools makes the arguments very difficult. In fact some of the results that we obtain formally will motivate concrete descriptions of generators. If you still don't want to use the long exact sequence, try looking up a proof it and see if you can apply the exactness arguments to this specific case.

Specifically I'm using singular (co-)homology, where we consider the space $C(\Delta^n, X)$ of continuous functions from the $n$-simplex to $X$ and then form the chain groups $C_n(X)$ as the free abelian group on this set. This means that to construct a homomorphism $\varphi \colon C_n(X) \to G$ it is enough to define a plain function $C(\Delta^n, X)\to G$. The relative chains are defined as $C_n(X, A) = C_n(X)/C_n(A)$ and have a basis consisting of cosets $\{f + C_n(A)\}$ where the image of $f$ is not contained in $A$. 
Let $c_t\colon \Delta^0\cong\{t\} \to I$ be the inclusion for $t\in I$. Then as a $0$-chain $c_t$ generates $H_0(I)\cong \mathbb{Z}$ for any $t\in I$, and the chains $c_0$ and $c_1$ generate $H_0(\partial I) \cong \mathbb{Z} \oplus \mathbb{Z}$.
Let $\alpha\colon \Delta^1 \to I$ be the linear simplex which sends $e_1$ to $0$ and $e_2$ to $1$. Then $\partial\alpha = c_1 - c_0$ which is supported in $\partial I$, so $\alpha$ is a relative cycle. Then we can see from the long exact sequence of the pair $(I, \partial I)$ that the cohomology class of $\alpha$ is a generator of $H_1(I, \partial I)$ :
$$ 0\to H_1(I, \partial I) \stackrel{\partial}{\to} H_0(\partial I) \to H_0(I) \to 0$$
The kernel of $H_0(\partial I) \to H_0(I)$ is the anti-diagonal and $\partial\alpha$ generates this subgroup, so by exactness $\alpha$ is a generator.
Now since $(I, \partial I)$ is a "good pair" we get an isomorphism of homology groups 
$$ q_*\colon H_1(I, \partial I) \cong H_1(S^1)  $$
induced by taking the quotient by $\partial I$. Along this isomorphism we explicitly see our class $\alpha$ gets mapped to a cycle which wraps once around the circle, which is the usual generator of this group.
Dualizing:
Now our relative cochain groups are
$$ C^n (X, A) = Hom(C_n(X, A), \mathbb{Z})$$ 
By the definition of relative singular chains, we can construct a relative cochain by giving a function $\varphi\colon C(\Delta^n, X) \to \mathbb{Z}$ which vanishes on $C(\Delta^n, A)$. Then $\delta$ is the formal dual of $\partial$, and we get $H^n(X, A)$ as relative co-cycles modulo relative co-boundaries.
There is a natural Kronecker pairing
$$H^n(X, A) \times H_n(X, A) \to \mathbb{Z} $$
inducing a homomorphism $H^n(X, A) \to Hom(H_n(X, A), \mathbb{Z})$. Remarkably this homomorphism is surjective, but by the Universal Coefficient Theorem its kernel is $Ext(H_{i-1}(X, A), \mathbb{Z})$. When these $Ext$ groups vanish (which they do in our case) then cohomology is nothing more than a homomorphism from homology to the coefficient group.
A generator of $H^0(I)\cong \mathbb{Z}$ is given by the function $\kappa\colon C(\Delta^0, I) \to \mathbb{Z}$ with constant value $1$. The group $H^0(\partial I)\cong \mathbb{Z}\oplus \mathbb{Z}$ is generated by $c_0^* = \kappa|_0$ and $c_1^*\kappa|_1$, so that $c_i^*(c_i) = 1$. Now our cohomology exact sequence is as you describe
$$0 \to H^0(I) \to H^0(\partial I) \stackrel{\delta}{\to} H^1(I, \partial I) \to 0 $$
Since the image of the first map is the diagonal we can use exactness to deduce that $H^1(I, \partial I) \cong (\mathbb{Z}\oplus \mathbb{Z})/diag \cong \mathbb{Z}$. Moreover we can also say that it is generated by $\delta c_1^* = \delta(0, 1)$, but it would be nice if we had a better description of this cohomology class. Remember by definition
$$ (\delta c_1^*)(c) = c_1^*(\partial c) $$
If $c$ is a relative $1$-cycle, then our computation of the homology groups implies that $c$ is homologous to $k\alpha$ for some $k$, and therefore $(\delta c_1^*)(c) = c_1^*(k c_1 - k c_0) = k$. In fact $\delta c_1^*$ is the homomorphism $H_1(I, \partial I) \to \mathbb{Z}$ sending the class of $\alpha$ to $1$. 
You can define it explicitly as a relative cochain by defining it on elementary chains $f\in C(\Delta^1, I)$ by saying that $\alpha^*(f) = 1$ if $f = \alpha$ and $0$ otherwise. $\alpha^*$ is a relative cocycle because for any relative $2$-chain $c$ we have $ (\delta \alpha^*)(c) = \alpha^*(\partial c) $, so if $\alpha^*(\partial c) \neq 0$ then $\partial c = k\alpha \in C_1(I, \partial I)$ for some $k$. However we know that $\alpha$ and its multiples are not boundaries since it generates $H_1(I, \partial I) \cong \mathbb{Z}$! Therefore $\alpha^*$ is a cocycle, and since $\alpha^*(\alpha )= 1$ its cohomology class agrees with $\delta c_1^*$.
Back to the circle question, $q^*\colon H^1(S^1) \cong H^1(I, \partial I)$. Then since $q_*(\alpha)$ is the usual generator of $S^1$, the dual of this generator pulls back to $\alpha^*$.
