I always read the snake lemma with short exact sequences: \begin{eqnarray*} &&\qquad M_1\to M_2\to M_3\to0\\ &&\qquad\ \downarrow\qquad\downarrow\qquad\ \downarrow\\ &&0\to N_1\to N_2\to N_3 \end{eqnarray*}

But does it hold with longer exact sequences? \begin{eqnarray*} &&\qquad M_1\to M_2\to M_3\to M_4\to M_5\to\cdots\to M_n\to0\\ &&\qquad\ \downarrow\qquad\downarrow\qquad\ \downarrow\qquad\downarrow\qquad\ \downarrow\qquad\qquad\,\ \downarrow\\ &&\ 0\to N_1\to N_2\to N_3\to\ N_4\to\ N_5\to\cdots\to N_n \end{eqnarray*}

Thanks in advance.

  • $\begingroup$ Try diagram chasing a la the proof of the snake lemma. $\endgroup$ – Avi Steiner May 30 '14 at 2:21
  • $\begingroup$ So the answer is yes? ;) $\endgroup$ – joaopa May 30 '14 at 2:22
  • $\begingroup$ Dunno. But trying the argument should give you a good idea whether or not it is. $\endgroup$ – Avi Steiner May 30 '14 at 2:23
  • $\begingroup$ I edit my post to explicit the arrows $\endgroup$ – joaopa May 30 '14 at 6:06

Nope. Given any short exact sequence $0\to A\to B\to C\to 0$ mapping to $0\to A'\to B'\to C'\to 0$ you can extend to $0\to 0\to A\to B\to C\to 0$ mapping to $0\to D\to A'\oplus D\to B'\to C'\to 0$. Given $c\in C$ in the kernel of $C\to C'$, as usual we can map it to something in $A'$, but it won't be in the image of $D$ unless it's $0$.


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