Showing the connecting homomorphism is well defined (cohomology) I want to show that the connecting homomorphism $d^*:H^k(\mathcal{C})\to H^{k+1}(\mathcal{A})$ as defined in Loring Tu's "Introduction to Manifolds" is well defined.
Suppose for each $k$, $0\rightarrow A^k\xrightarrow{\mathit{i_k}} B^k\xrightarrow{\mathit{j_k}} C^k\rightarrow0$ is a short exact sequence of vector spaces (will omit subscript $k$ from now on). Given a short exact sequence, he constructs $d^*$ as follows. Consider the short sequences in dimensions $k$ and $k+1$.
Start with $[c]\in H^k(\mathcal{C})$. Since $j:B^k\to C^k$ is onto, there is a $b\in B^k$ such that $j(b)=c$. Then $db\in B^{k+1}$ is in $\ker j$ because $jdb=djb=dc=0$, as $c$ is a cocycle. By the exactness of the sequence in degree $k+1$, $\ker j= \text{im }i$, so $db=i(a)$ for some $a\in A^{k+1}$, and this $a$ is unique since $i$ is injective, which also implies $da=0$, since $i(da)=d(ia)=ddb=0$. Hence $a$ is a cocycle and defines a cohomology class $[a]$, and we set $d^*[c]=[a]$.
To show this map is well defined, I want to show that this definition is independent of the choice of $b$, and independent of the choice of representative of $[c]$. The former, I have successfully shown. But I do not see how to show that the result is independent of the choice of representative. I think this comes down to a lack of understanding of what it means to be in a cohomology class.
 A: So suppose that $c$ and $c'$ are two alternative representatives of the class $[c]$. We follow your construction through for both $c$ and $c'$. Concretely, this means we find $b$ and $b'$ such that $j(b) = c$ and $j(b') = c'$, and we find $a$ and $a'$ such that $i(a) = d(b) $and $i(a') = d(b')$.
Our task is to show that $a$ and $a'$ represent the same class in $H^{k+1}(\mathcal A)$. In other words, our task is to find a $w \in A^k$ such that $d(w) = a' - a$.
Let's construct this $w$.
First, the fact that $c$ and $c'$ represent the same class in $H^k (\mathcal C)$ means that there exists a $u \in C^{k-1}$ such that $d(u) = c' - c$.
Since $j : B^{k-1} \to C^{k-1}$ is surjective, there exists a $v \in B^{k-1}$ such that $j(v) = u$.
Now consider the element $z := b' - b - d(v)$. Notice that
$$
j(z) =  j(b') - j(b) - j(d(v)) = j(b') - j(b) - d(j(v)) =  j(b') - j(b) - d(u) \\ = c' - c - (c' - c) = 0
$$
So $z$ is in the kernel of $j$. By exactness, $z$ must also be in the image of $i$, i.e. there must exist a $w \in A^k$ such that $i(w) = z$.
Now observe that
$$ i(d(w) - (a' - a)) = d(i(w)) - (i(a') - i(a)) = d(z) - (i(a') - i(a)) \\ = (d(b') - d(b) - d(d(v))) - (d(b') - d(b)) = 0. $$
Since $i$ is injective, we conclude that
$$ d(w) = a' - a.$$
Thus we've constructed the $w$ that we set out to construct! This $w$ is evidence to the fact that $[a] = [a']$ in $H^{k+1}(\mathcal A)$.
