# Connecting morphism in an abelian category

I'm trying to understand how one gets the long exact sequence in homology from a short exact sequence of chain complexes in an arbitrary abelian category.

So far I have the commutative diagram below, with all rows exact:

We examine the homologies

Where $\bar H$ is the dual object of the homology, which is isomorphic to it.

Now the author of the book I'm following (Osborne's Basic Homological Algebra, p.231) says that the connecting morphism is defined as the homology map induced by $f_2$: $$(f_2)_\ast =\delta:\bar H\rightarrow H$$ I do not follow. How do we know such an arrow exists? In what precise sense is it induced by $f_2$?

Added: I would like a detailed explanation; "Just apply the universal properties" is not the answer I'm looking for.

• Have you heard of the Snake Lemma? – Hayden Jan 19 '15 at 23:15
• @Hayden yep. Unless I'm mistaken, I'm asking about the connecting morphism given by the snake lemma. My question comes from the process of proving it in the book I'm following, so I don't want to use it. – user153312 Jan 19 '15 at 23:18
• Ah, I see, just thought I'd mention it since you never explicitly said "snake lemma" (though your diagram is exactly what it applies to). – Hayden Jan 19 '15 at 23:19
• I don't know a way to make $(f_2)_*$ the image of $f_2$ under an operation $(-)_*$, but an explicit (and element-free) definition of the connecting homomorphism is given here. – tcamps Jan 20 '15 at 3:49
• @tcamps I don't understand "This then induces a map on quotients $B^{\prime \prime}/A \rightarrow A^\prime /\operatorname{Im}a$ which is precisely the desired map $\operatorname{Ker}c\rightarrow \operatorname{Coker}a.$" Explicitly, how is this map induced? I wasn't able to figure it out by myself. Also, it is $A^\prime /\operatorname{Im}a$ and not $A^\prime /\operatorname{Im}A$, right? – user153312 Jan 20 '15 at 13:11