How do you get well-definedness in long exact sequence connecting homomorphism? Let $0 \to K_{\bullet}' \xrightarrow{a} K_{\bullet} \xrightarrow{b} K_{\bullet}'' \to 0$ be a short exact sequence of modules.  There exists a connecting map $c: H_{n}'' \to H_{n-1}'$ derived as follows:
Let $\bar{x} \in H_n''$ and so in particular $x \in \text{im} \ b_n$ by exactness of the complexes, so let $y \in b_n^{-1}(x)$.  But $d b_n = b_{n-1} d$ so that $dx = b_{n-1}dy = 0$.  We now need $\bar{z} \in H_{n}'$ to map to $y$. Since $dy \in \ker b_{n-1}$ by exactness of the complexes we have $dy \in \text{im} \ a_{n-1}$ so that $a_{n-1} z = dy$ for some $z \in K_{n-1}'$.
That's where I'm stuck in the mud.  
 A: So far, so good.  Now, 
$$
a_{n-2}(dz) = d(a_{n-1}(z)) = d(dy)=0
$$
and as $a_{n-2}$ is injective we have $dz=0$.  Thus $z$ is a cycle in $K_{n-1}^{'}$ and thus represents an element in $H_{n-1}^{'}$.  Hence we may define $c: H_n^{''} \to H_{n-1}^{'}$ by $c(\bar{x}) = \bar{z}$.  Now argue that this is well-defined and a homomorphism. 
A: I find it much easier to make this sort of argument using all preimages rather than just choosing one.
That is, in  first step of taking a preimage under $b_n$, rather than just working with the single preimage $y$, I instead take the entire class $y + a_n(K'_n)$. One can view this as an element of the quotient module $K_n / a_n(K'_n)$.
Taking the differential goes to $dy + da_n(K'_n) = dy + a_{n-1} d(K'_n)$.
Then, taking the preimage under $a_{n-1}$ gives $z + d(K'_n)$.
By doing things this way, I find it easier to see:


*

*that the result is linear in $x$

*that the chase yields a unique element of $K'_{n-1} / d(K'_n)$


and we thus get a well-defined homomorphism $\ker\{ K''_n \xrightarrow{d} K''_{n-1} \} \to K'_{n-1}/d(K'_n)$.
