Weibel Exercise 2.2.3 is as follows:
Let $P_\bullet$ be a complex of projectives with $P_i=0 \; \forall i<0$ then a map $\epsilon : P_0 \to M$ giving a resolution for $M$ is the same thing as a chain map $\epsilon : P_\bullet \to M_\bullet$, where $M_\bullet$ is the chain complex with $M$ concentrated in degree $0$.
But I am clearly misunderstanding this...
Clearly if we have such a map; $$ \to P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{\epsilon_0} M \to 0$$
Making a projective resolution for $M$, then we have $ker(\epsilon_0)=Im(d_1)$ so commutative ladder diagram:
$$ \begin{array}{cccccc} ... \xrightarrow{d_2} & P_{1} & \xrightarrow{d_1} & P_0 & \xrightarrow{0} & 0 & \xrightarrow{0} ...\\ & \big\downarrow & & \; \big\downarrow \epsilon & & \big\downarrow \\ ... \xrightarrow{0} & 0 & \xrightarrow{0} & M & \xrightarrow{0} & 0 & \xrightarrow{0} ... \end{array} $$
.
However I cannot see how the two things are the same. We have no guarantee of exactness in the complex $P_\bullet$ for one. As well as no guarantee of equality $Im(d_1) = ker(\epsilon_0)$ and surjectivity of $\epsilon_0$.
Are them some missing conditions or have I misunderstood the statement?
This is an error in Weibel, included in the list of errata here: http://www.math.umd.edu/~jmr/602/bookerrors.pdf
"chain map" should be replaced by "quasi-isomorphism", which will fix the problem you observed.