# Projective Resolution and Homology

Let $$A$$ be an abelian category with enough projectives. We construct a projective resolution of an object C as follows: since $$A$$ has enough projectives we have an epimorphism from a projective object $$P^0$$ onto C. Let $$ΩΑ$$ denote its kernel. Given $$ΩΑ^{i}$$ take an epimorphism $$P^i$$-->$$ΩΑ^{i}$$ and let $$ΩΑ^{i+1}$$ denote its kernel. Concaneting these short exact sequences together we obtain a projective resolution of C. I am having trouble to understand why the "glued" sequence is exact at every position.

• Sure! A sequence A-->B-->C is exact at B if gof=0 and Imf is isomorphic with Kerg. In category theory, kernel of a morphism f:X-->Y is defined to be the pull back of a certain diagram, which is an object with a universal property as you noted. – no name Jan 18 at 18:26

As you say, given $$A\in\text{Ob}(\mathcal{A})$$, where $$\mathcal{A}$$ is an abelian category with enough projectives, you construct a resolution of $$A$$ by considering short exact sequences of the form $$0 \to \Omega^{i+1}(A)\to P^{i}\to \Omega^{i}(A)\to 0,$$ where $$P^{i}$$ is projective and $$\Omega^{0}(A)=A$$, and gluing them together: the map $$\phi^{i}:P^{i}\to P^{i-1}$$ is the composition $$P^{i}\to \Omega^{i}(A)\to P^{i-1}.$$ The kernel of this map is $$\Omega^{i+1}(A)$$ and the image is $$\Omega^{i}(A)$$ by construction. Therefore if you have the sequence $$P^{i+1}\xrightarrow{\phi^{i+1}}P^{i}\xrightarrow{\phi^{i}} P^{I}$$ you have $$\text{im}(\phi^{i+1}) = \Omega^{i+1}(A)$$ and $$\text{ker}(\phi^{i})=\Omega^{i+1}(A)$$, so the sequence is exact. In particular, the resolution is exact at each point.