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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?

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    $\begingroup$ You are right!just adding that the chain map is quasi-isomorphism. $\endgroup$
    – Jian
    May 21, 2018 at 12:17
  • $\begingroup$ @Sky this works, thanks!! $\endgroup$
    – SEWillB
    May 21, 2018 at 12:51
  • $\begingroup$ What does "projective" do? $\endgroup$
    – Ryze
    Jun 15, 2020 at 5:05

1 Answer 1

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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.

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