Let $0\to K'\to K\to K''\to 0$ be a short exact sequence of chain maps and suppose the boundary map is going from $K'_n\to K'_{n-1}$ and similarly for $K,K''$. Then this induces homology long exact sequence.
$\textbf{Q:}$ Suppose one knows naturality of homology functor and existence of long exact sequence. Then connecting homomorphism $\partial_\star:H_\star(K'')\to H_{\star-1}(K')$ is determined up to a sign. What is the meaning of the question? Or why do I see there is a sign? The connecting homomorphism is usually defined by a "switchback" map.(i.e. pull back along $K\to K''$ to a represenative and descend to the lower $K$ to see this defining a cycle in $K'$ which requires further descending.) What is the meaning of this $\pm 1$ here? If I have $\pm 1$ issue, then I would certainly have trouble for naturality.
Ref. Dold Algebraic Topology pg 23. 2.16 Exercise.2
Best Answer
Let $0\to A\to B\to C \to 0$ be a SES of chain complexes.
Then we're given that there is a LES of homology, with $f_*: H_n(A)\to H_n(B)$, $g_*:H_n(B)\to H_n(C)$, and $\partial_*:H_n(C)\to H_{n-1}(A)$, all of which are natural (in the sense that they represent natural transformations of functors on SESs).
Now note that $-\partial_*$ is still natural, and the LES is still exact if we replace $\partial_*$ with $-\partial_*$.
That's all that the book is trying to say.
The goal of this exercise is to show that these two possibilities are the only possible options for natural maps $H_n(C)\to H_{n-1}(A)$ that still make the LES exact. However, you don't seem to be asking how to show the exercise, just where the $-1$ comes from, so I'll leave it here.