I was reading about projective resolution and I came across this following definition:
A projective resolution of an R-module M is a complex X together with an R-linear map $ X_0 \xrightarrow \epsilon M$ such that the sequence
$.... \rightarrow X_n \xrightarrow {d_{n}} X_{n-1}......X_1\xrightarrow {d_{1}} X_0 \xrightarrow \epsilon M\rightarrow 0 $ is exact and all the $X_i$ are $R$-projective.
Clearly X has a property that $H_n(X) =0$ , $n>1$, and $H_0(X) \cong M$, where $H_n(X)= \frac{ker(d_n)}{im(d_{n+1})}$
I couldn't understand the last sentence, i.e how can we prove $H_0(X) \cong M$. From the definition of homology we can see $H_0(X)= \frac{ker(d_0)}{im(d_1)}$. But the sequence is exact, then that must imply that $H_0(X)$ is equal to 0. And secondly what about $H_1(X)$, shouldn't it be 0 also.
It will be great if someone can kindly clear the doubts I am having .
By exactness, $P_1\to P_0 \to M\to 0$ gives you that $P_0/\text{im}(P_1\to P_0)\cong M$. The complex you are using to compute homology is slightly different, as you don't have $M$ there. Instead, the end is $P_1\to P_0\to 0$, so $P_0$ is the $0$th kernel, and hence $H_0(P_*)=P_0/\text{im}(P_1\to P_0)$.