I have been reading MacLane's Book on Homology and im reading the part about resolutions of a module where he defines it like this
A resolution $(X,\epsilon )$ of C is an exact sequence $\rightarrow X_n \rightarrow X_{n-1} \rightarrow ... \rightarrow X_1 \rightarrow X_0 \rightarrow C \rightarrow 0$, where $ \epsilon :X_0 \rightarrow C$.
Then he goes on to say ,
That is a complex $(X,\epsilon)$ over C with $H_n(X)=0$, $n >0$ and $\epsilon : H_0(X) \cong C$.
And this is the part i dont get , if the sequence was exact shouldnt the Homology groups be all trivial ? Or does he exact sequence end at $X_0$? Thanks in advance.
By definition, the complex \begin{equation}\tag{1} \dotsb \rightarrow X_n \rightarrow \dotsb \rightarrow X_1 \rightarrow X_0 \rightarrow C \rightarrow 0 \end{equation} is exact, but the complex \begin{equation}\tag{2} \dotsb \rightarrow X_n \rightarrow \dotsb \rightarrow X_1 \rightarrow X_0 \rightarrow 0 \end{equation} is actually not. Indeed, given that (1) is exact, we find that all the homology groups of (2) but the $H_0$-group are trivial, and for the $H_0$-group we have $$ H_0(X_\bullet) = \dfrac{\ker(X_0 \rightarrow 0)}{\operatorname{im}(X_1 \rightarrow X_0)} = \dfrac{X_0}{\operatorname{im}(X_1 \rightarrow X_0)} = \operatorname{coker}{(X_1 \rightarrow X_0)} \simeq C. $$