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Title is pretty straightforward, it is almost a duplicate of Example of Short exact Sequence of chain complexes but I was unable to understand the answers. I have tried many complexes, and unable to find one where the connecting homomorphism. If possible, please provide a simple example which doesn't involve topology. Hopefully something with complexes containing $\mathbb{Z}$ only.

Shean
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  • What is too complicated with paul garrett’s answer to the linked thread? There’s no topology in there are the spaces involved are fairly small. – Aphelli Feb 02 '23 at 20:34
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    While the title is straightforward, your question should be understandable without its title. And it isn't. – Mariano Suárez-Álvarez Feb 03 '23 at 02:01
  • In any case pick any non split exact sequence 0\to A\to B\to C\to 0, resolve the two endpoints A and C projectively, use the horseshoe lemma to obtain a projective resolution of the middle term B. Now you have an exact sequence of complexes. Apply hom(-,A) to everything to obtain a short exact sequence of complexes. Its connecting morphism is not zero. – Mariano Suárez-Álvarez Feb 03 '23 at 02:04

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This one's pretty simple: $$ \begin{array}{ccccccccc} 0 & \to & 0 & \to & 0 & \to & 0 & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & 0 & \to & \mathbb{Z} & \to & \mathbb{Z} & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & \mathbb{Z} & \to & \mathbb{Z} & \to & 0 & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & 0 & \to & 0 & \to & 0 & \to & 0 \\ \end{array} $$ where all maps $\mathbb{Z} \to \mathbb{Z}$ are the identity.

John Palmieri
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    Indeed, it cannot get simpler than drawing the smallest SES of complexes of free modules that supports the chase in the snake lemma. :) – Pedro Feb 03 '23 at 21:42
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$\DeclareMathOperator{\Ext}{Ext} \newcommand{\Z}{\mathbb Z}$Let me explain why Mariano's approach works. Suppose indeed that $\xi : 0 \to A \to B \to C\to 0$ is a non-split short exact sequence of modules over some ring $R$, meaning it does not represent the zero class in $\Ext^1(C,A)$.

Consider resolutions $P'\to A$ and $P''\to C$, and then build a resolution $P$ of $B$ of the form $P'\oplus P''$ (but a priori a differential that adds a morphism at the top right, as in the Horshoe Lemma).

If you apply $\hom(-,A)$ to $P'\to P\to P''$ and take homology, what you get is the long exact sequence

$$\hom(C,A) \to \hom(B,A) \to \hom(A,A) \to \Ext^1(C,A) \to \cdots$$

One can show that the image of the identity $A\to A$ in $\hom(A,A)$ is sent to the class of $\xi$ in $\Ext^1(C,A)$, which is non-zero, so the connecting morphim $\hom(A,A) \longrightarrow \Ext^1(C,A)$, which is given by right multiplication by $[\xi]$, is not zero.

For a concrete example, let $A = C = \mathbb Z/2$ and $B = \mathbb Z/4$ over $R = \mathbb Z$. Then any short exact sequence with these modules is not split, and the connecting morphism is not zero.

Add. If you follow this construction in this case, you will get the following SES of complexes

$$ \begin{array}{ccccccccc} 0 & \to & \Z & \to & \Z^2 & \to & \Z & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & \Z & \to & \Z^2 & \to & \Z & \to & 0 \end{array} $$

where the horizontal maps are $i_1$ and $\pi_2$, the two vertical maps $\Z\to \Z$ are multiplication by $2$, and the third vertical map is given by the matrix

$$\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$$

If you do $\hom_\Z(-,\Z/2)$ and follow the definition of the chase in the snake lemma, you will see that the connecting morphism $$\delta : \hom_{\Z}(\Z/2,\Z/2)\cong \mathbb Z/2 \to \mathbb Z/2 \cong \Ext_\Z^1(\Z/2,\Z/2)$$ is the identity. Indeed, the vertical maps $\Z/2\to \Z/2$ are zero, the map $(\Z/2)^2 \to (\Z/2)^2$ is the nilpotent map
$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ and this is all you need to conclude.

Pedro
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