What does "modulo" mean in a chain complex context?

I am trying to self-learn category theory, and right now I want to get down the idea of what I derived functor is (with the end goal of cohomology). My current main source is Wikipedia, but on the page for derived functors they say "But we can compute its cohomology at the $i$-th spot (the kernel of the map from $F(I^i)$ modulo the image of the map to $F(I^i)$)".

On the linked page for cohomology, they also use the idea of modulo without defining it. I'm obviously well acquainted with factor groups and other such notions in critical examples of categories, but not in all generality. How is it done?

My guess is that they somehow show that up to isomorphism the exact sequence

$$0 \to H\to G\to G/H\to 0$$

has a unique solution $G/H$ and that is out factor object, but I can't seem to be able to prove the solution is unique.

Solution 1:

The set up from the wikipedia page your looking at is a chain complex starting at $0$, which is a sequence of homomorphisms of abelian groups $$ 0\xrightarrow{g_0}F(I^0)\xrightarrow{g_1}F(I^1)\xrightarrow{g_2}\cdots $$ such that image of each homomorphism is included in the kernel of the next, i.e. $\operatorname{im}(g_i)\leq\ker(g_{i+1})$.

So, the map from $F(I^i)$ is the map I've called $g_{i+1}$, while the map to $F(I^i)$ is $g_i$. Therefore, the sentence "the kernel of the map from $F(I^i)$ modulo the image of the map to $F(I^i)$" means the quotient group $\ker(g_{i+1})/\operatorname{im}(g_i)$. Note that $\operatorname{im}(g_i)\leq\ker(g_{i+1})$, as we are in a chain complex, so this quotient makes sense.