Theorem : Given an exact sequence $$0 \longrightarrow A
\longrightarrow B \longrightarrow C \longrightarrow 0$$ of
chain/cochain exists a connecting homomorphism $\omega : H(C)
\longrightarrow H(A)$ of degree $1$ or $-1$ which induces a long exact
sequence in homology or cohomology.
The construction of $\omega$ is quite explicit, in fact $\omega[\gamma] = \alpha$ where $\alpha$ is a preimage of $\partial \beta$, where $\beta$ is the element such that $\beta \longmapsto \gamma \in C_n$.
Since in the proof one proves that doesn't depend on the choice of the $\gamma$ representing $[\gamma]$ and $\beta$ which goes to $\alpha$, could I conclude that whenever I find an explicit function, such that respect those requirements for specific elements, that function is $\omega$ indeed since in homology they are the same function ?
The question arises from the connecting homomorphism of the $Cf$ complex, where for the first time I noticed this kind of "uniqueness" to assert that in that scenario, the long exact sequence has $\omega = f_*$ as connecting homomorphism.
Any clarification or help would be appreciated.
Best Answer
What I believe you are asking is whether one can replace the connecting homomorphism $\partial : H_n(C) \to H_{n-1}(A)$ with some other homomorphism so that the sequence is still exact. If that's the question then yes, this is possible. For example, as noted in the comments and in this post, the homomorphism $-\partial : H_n(C) \to H_{n-1}(A)$ has the same kernel and same image as $\partial$, and so you could replace $\partial$ with $-\partial$ without affecting exactness.
But one could say more: if $h : H_n(C) \to H_n(C)$ is any automorphism that preserves $\text{kernel}(\partial)$ --- meaning that $h(\text{kernel}(\partial)) = \text{kernel}(\partial)$ --- then you could replace $\partial$ by $\partial \circ h$; and if in addition $k : H_{n-1}(A) \to H_{n-1}(A)$ is an automorphism that preserves $\text{image}(\partial)$ then you could replace $\partial$ by $k \circ h$; and you could put these together and replace $\partial$ by $k \circ \partial \circ h$. Each of these alterations keeps the sequence exact.
So, for example, additive inversion in any abelian group is an isomorphism that preserves all subgroups, and so replacing $\partial$ with $-\partial$ does not affect exactness of the sequence.
For a broader kind of example, let's consider the case where the homology groups in question are free abelian groups. Free abelian groups have lots of automorphisms, and so it might be quite easy to find lots of $h$'s and $k$'s. For a more specific example, suppose that $H_n(C) \approx \mathbb Z^\beta$ is a free abelian group of rank $\beta$ ($=$ the $n^\text{th}$ Betti number of $C$). It follows that $$\text{Aut}(H_n(C)) \approx \text{Aut}(\mathbb Z^\beta) \approx \text{GL}_\beta(\mathbb Z) $$ If in addition $H_{n-1}(Z)$ is free abelian then the subgroup $\text{kernel}(\partial)$ is a direct factor of some rank $r$, and by appropriate choice of basis there is a direct sum decomposition $H_n(C) \approx \mathbb Z^\beta \approx \mathbb Z^r \oplus \mathbb Z^{\beta-r}$ so that $\text{kernel}(\partial)$ is identified with the first direct summand. In this situation you can choose $h$ to be any invertible matrix of integers having all $0$'s in the upper right $r \times (\beta-r)$ block.
But there's a catch: one very important fact about the standard construction of the connecting homomorphism $\partial$ that you describe in your post is that it is a natural transformation in the category theoretic sense. As explained in this post, $-\partial$ is also a natural transformation. Undoubtedly if you choose $h$ and $k$ carelessly (rather than just choosing additive inversion) then you'll lose naturality.