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Nov 27, 2012 at 18:37 comment added Joseph Victor Good point, mt. These are clearly exact.
Nov 27, 2012 at 8:42 comment added M T $j$ has a kernel equal to $\operatorname{im} \iota$, so $j^{-1}(N)$ contains $\operatorname{im} \iota$ and is never zero. In the case you describe, the sequence splits.
Nov 27, 2012 at 3:26 comment added Joseph Victor I tried something like this but game up. I don't think these are still exact. Pick some 1-d $N$ which maps onto $k$, then $j^{-1}(N)=0$ but exactness, but $k\to 0$ is not injective. Am I missing something?
Nov 26, 2012 at 9:12 history answered M T CC BY-SA 3.0