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.
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 $0$.