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Let $R$ be commutative ring with 1. Suppose we have the following exact sequences of $R$-modules.

\begin{array}{ccccccccc} 0 & \rightarrow & M'& \xrightarrow{f} & M& \xrightarrow{q} & M''& \rightarrow& 0\\ & & & & \downarrow{b}& & \downarrow{c}& & \\ 0 & \rightarrow & N'& \xrightarrow{f'} & N& \xrightarrow{q'} & N''& \rightarrow& 0\\ \end{array} Suppose the diagram above commutes. Then we get a morphism $a:M'\rightarrow N'$ such that we get the following commutative diagram.

\begin{array}{ccccccccc} 0 & \rightarrow & M'& \xrightarrow{f} & M& \xrightarrow{q} & M''& \rightarrow& 0\\ & & \downarrow{a} & & \downarrow{b}& & \downarrow{c}& & \\ 0 & \rightarrow & N'& \xrightarrow{f'} & N& \xrightarrow{q'} & N''& \rightarrow& 0\\ \end{array} Then by snake lemma we get an exact sequence $$0\rightarrow Ker\,a\rightarrow Ker\,b\rightarrow Ker\,c\xrightarrow{\phi} Coker\,a\rightarrow Coker\,b\rightarrow Coker\,c\rightarrow 0.$$

However, I have a naive doubt. We can replace the second exact sequence by the images of the corresponding maps.

\begin{array}{ccccccccc} 0 & \rightarrow & M'& \xrightarrow{f} & M& \xrightarrow{q} & M''& \rightarrow& 0\\ & & \downarrow{a} & & \downarrow{b}& & \downarrow{c}& & \\ 0 & \rightarrow & im\,a & \xrightarrow{f'} & im\,b &\xrightarrow{q'} & im\,c& \rightarrow& 0\\ \end{array}

Applying snake lemma to above diagram, we get $$0\rightarrow Ker\,a\rightarrow Ker\,b\rightarrow Ker\,c\rightarrow 0$$.

This seems to mean that $\phi$ is the zero map. Am I making a mistake somewhere?

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    $\begingroup$ If we replace the second exact sequence by the images, then it might not be exact, or even well defined. $\endgroup$ – Crostul Feb 1 '18 at 13:25
  • $\begingroup$ @Crostul Why not well-defined? Since $f' \circ a = b \circ f$, applying $f'$ to an element in the image of $a$ should give an element in the image of $b$. Same argument for $q'$. $\endgroup$ – Y. Forman Feb 1 '18 at 13:50
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As Crostul said in the comment, we have no guarantee that the new second sequence will be exact.

For example, let's take the following diagram of $\mathbb{Z}$-modules (the unlabeled maps are reduction maps):

\begin{array}{ccccccccc} 0 & \rightarrow & \mathbb{Z}& \xrightarrow{\cdot 4} & \mathbb{Z}& \xrightarrow{} & \mathbb{Z}/4& \rightarrow& 0\\ & & \downarrow{\cdot 2} & & \downarrow{id}& & \downarrow{}& & \\ 0 & \rightarrow & \mathbb{Z}& \xrightarrow{\cdot 2} & \mathbb{Z}& \xrightarrow{} & \mathbb{Z}/2& \rightarrow& 0\\ \end{array}

Now the right and center downward maps are surjective; the left downward map has image $2\mathbb{Z}$; which we can write as $\mathbb{Z}$ and interpret the rightward $\cdot 2$ map as $\cdot 4$. Thus the sequence of images is

$$ 0 \rightarrow \mathbb{Z} \xrightarrow{\cdot 4} \mathbb{Z} \rightarrow \mathbb{Z}/2 \rightarrow 0$$

which is not exact.

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  • $\begingroup$ Ah, thank you. As I see, the issue seems to me that the surjectivity of centre and right downwards arrow does not give surjectivity of left downward arrow. $\endgroup$ – user52991 Feb 1 '18 at 14:21

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