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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.

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  • $\begingroup$ 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. $\endgroup$ – no name Jan 18 at 18:26
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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.

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