This proposition is as follows: Let $$ 0\rightarrow L\rightarrow M\rightarrow N\rightarrow 0 $$ be a short exact sequence of R-modules. Let the following be projective resolutions of L and N respectively: $$ \cdots\rightarrow P_{1}\rightarrow P_{0}\rightarrow L\rightarrow 0 $$ $$ \cdots\rightarrow \bar{P}_{1}\rightarrow \bar{P}_{0}\rightarrow N\rightarrow 0 $$ The proposition claims that there is a projective resolution for M: $$ \cdots\rightarrow P_{1}\oplus\bar{P}_{1}\rightarrow P_{0}\oplus\bar{P}_{0}\rightarrow M\rightarrow 0 $$ that commutes with all the above exact sequences, as well as the obvious ones given by $$ 0\rightarrow P_n\rightarrow P_n\oplus\bar{P}_{n}\rightarrow\bar{P}_{n}\rightarrow 0 $$

The book constructs the maps in the projective resolution for M as follows: For the first map, first lift the map $\bar{P}_{0}\rightarrow N$ to $\Phi:\bar{P}_{0}\rightarrow M$ (this is possible because $M\rightarrow N\rightarrow 0$ is surjective and $\bar{P}_{0}$ is projective). Then, compose $P_{0}\rightarrow L$ and $L\rightarrow M$ to get a map $\psi:P_{0}\rightarrow M$, and define $\phi:P_{0}\oplus \bar{P}_{0}\rightarrow M$ as follows: $$ \phi(a,b)=\psi(a)+\Phi(b) $$ It is not hard to see that this makes the diagram commute. However, my problem is that I cannot see how the sequence given by the maps is exact at $P_{0}\oplus \bar{P}_{0}$.

I want to prove that if $\phi(a,b)\in ker(\phi)$, then $\phi(a,b)\in im(\alpha)$ where $\alpha:P_{1}\oplus \bar{P}_{1}\rightarrow P_{0}\oplus \bar{P}_{0}$. That means I need $\psi(a)$ to be zero so that $a$ is in the image of $P_{1}\rightarrow P_{0}$. However, there does not seem to be any reason for this to be true.

In particular, I have the following seeming counterexample. Suppose the initial exact sequence was $$ 0\rightarrow n\mathbb{Z}\rightarrow \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\rightarrow 0 $$ and the projective resolutions are: $$ \cdots\rightarrow 0\rightarrow\mathbb{Z}\xrightarrow{n}n\mathbb{Z}\rightarrow 0 $$ $$ \cdots\rightarrow\mathbb{Z}\xrightarrow{\alpha}\mathbb{Z}\oplus\mathbb{Z}\xrightarrow{\phi}\mathbb{Z}\rightarrow 0 $$ $$ \cdots\rightarrow\mathbb{Z}\xrightarrow{n}\mathbb{Z}\xrightarrow{n\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\rightarrow 0 $$ Then we have $$ \phi(a,b)=an+b $$ $$ \alpha(c)=(0,nc) $$ by commutativity of the whole diagram (which unfortunately I do not know how to draw). However, $(-1,n)\in ker(\phi)$ but $(-1,n)\not\in im(\alpha)$. This seems to contradict exactness.

It would be great if someone could point out the flaws in my argument. Thanks very much in advance!


What you're missing is that the map $P_1\oplus \bar{P}_1\to P_0\oplus \bar{P}_0$ should not be just defined to be the direct sum of the original maps $P_1\to P_0$ and $\bar{P}_1\to\bar{P}_0$. Instead, let $L_1$ denote the kernel of $P_0\to L$, let $N_1$ denote the kernel of $\bar{P}_0\to N$, and let $M_1$ denote the kernel of your map $\phi:P_0\oplus\bar{P}_0\to M$. Then there is a canonical short exact sequence $$0\to L_1\to M_1\to N_1\to 0$$ (exercise: construct the maps in this sequence and check that it is exact). Furthermore, $$\cdots\rightarrow P_{2}\rightarrow P_{1}\rightarrow L_1\rightarrow 0$$ and $$\cdots\rightarrow \bar{P}_{2}\rightarrow \bar{P}_{1}\rightarrow N_1\rightarrow 0$$ are projective resolutions of $L_1$ and $N_1$. So we can use the exact same procedure we used to construct $\phi$ to construct a surjection $\phi_1:P_1\oplus\bar{P}_1\to M_1$, which we then compose with the inclusion $M_1\to P_0\oplus\bar{P}_0$ to get the map $P_1\oplus \bar{P}_1\to P_0\oplus \bar{P}_0$, whose image is manifestly the kernel of $\phi$.

More generally, all of the maps $P_{n+1}\oplus \bar{P}_{n+1}\to P_n\oplus \bar{P}_n$ in the resolution of $M$ are constructed by iterating this process. These maps make the big diagram commute for the same reason that you saw that $\phi$ made the diagram commute.

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  • $\begingroup$ Thank you very much for clearing this up! I foolishly assumed that the map should be the canonical one, when in fact I should have simply repeated the initial process. Your answer is very clear and helpful. It is a pity I do not have enough reputation to change the vote count yet. $\endgroup$ – Houndoom Oct 13 '15 at 11:22

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