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Consider the following exact sequences:

$$ 0 \to A \to B$$ and $$B \to C \to 0$$

The first is exact at $A$ iff the kernel of the map from $A$ to $B$ is $\{ 0 \}$. The second is exact at $C$ iff the image of the map from $B$ to $C$ is all of $C$. Couldn't we say that this sequence is also exact at $C$ if the kernel of the map from $C$ to $0$ is $\text{ker}( \ 0 \to B$) because $B \to C \to 0$ is also $0 \to B \to C \to 0$ and $0 \to A \to B$ is also $0 \to A \to B \to 0$?

smartguy
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  • I think the misconception comes from the habit (that I do not share nor support) that some authors write $A\to B\to C$ as shorthand for $0\to A\to B\to C\to 0$. Thank you for exhibiting the need not to be so sloppy. – Hagen von Eitzen Jun 10 '13 at 18:14

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The kernel of the map from $C$ to $0$ is (by definition) a subobject of $C$, which $\ker(0\to B)$ is not. Moreover, the kernel of the map from $C$ to $0$ is always all of $C$ - exactness at $C$ is a question of what the image of the map $B\to C$ is.

You can't just "add" extra $\to 0$ and $0\to $ at the ends of an exact sequence - it won't always be exact. That is, just because $0\to A\to B$ is exact at $A$ doesn't mean that $0\to A\to B\to 0$ is also an exact sequence.

Zev Chonoles
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