# A consequence of the Snake lemma

This question is from Aluffi's book. I'm still trying my best to understand, "Snake's lemma" and consequences of that. I would be much appreciated for any suggestions to solving this problem. Thank you.

The snake lemma tells you that if you have a commutative diagram with exact rows $$\require{AMScd} \begin{CD} 0@>>> M_2 @>\alpha>> M_1 @>\beta>> M_0 @>>>0\\ @. @VVg_2V @VVg_1V @VVg_0V\\ 0@>>> N_2 @>\alpha'>> N_1 @>\beta'>> N_0 @>>>0 \end{CD}$$ then there is an induced exact sequence $$0\to\ker g_2\to \ker g_1\to \ker g_0\to\operatorname{coker}g_2\to\operatorname{coker}g_1\to\operatorname{coker}g_0\to 0.$$ If you know that each sequence $0\to L_i\xrightarrow{g_i} M_i\xrightarrow{f_i} N_i\to 0$ is exact, then you know that $g_i : L_i\to M_i$ maps $L_i$ isomorphically to $\ker f_i$, and that $\operatorname{coker}f_i\cong 0$. So, thinking of $L_i$ as $\ker f_i$, and applying the snake lemma to the rightmost two rows of the diagram (as in my diagram above), you have an exact sequence $$0\to L_2\to L_1\to L_0\to 0\to 0\to 0\to 0,$$ so the left column is exact, as long as the maps agree with the maps induced by the snake lemma. However, this will be forced: the maps occuring in the snake lemma come from the universal properties of kernels and cokernels, which force the maps to be unique. For example: you have a map $L_1\to M_1\to M_0\to N_0$, which agrees with the map $L_1\to M_1\to N_1\to N_0$, which is $0$ by exactness of $L_1\to M_1\to N_1$. This means that there is a unique map $\ell : L_1\to\ker(g_0) = L_0$ making the diagram $$\begin{CD} L_1 @>\ell>> L_0 @>>> 0\\ @VVf_1V @VVf_0V @VVV\\ M_1 @>\beta>> M_0 @>g_0>> N_0 \end{CD}$$ commute. Because your original map also made the diagram commute, uniqueness forces the two to agree. Similar arguments show the other maps must also be the same maps as in the snake lemma, and a dual argument with cokernels shows the result when the two leftmost columns are exact rather than the rightmost two.