Consider a commutative diagram in an abelian category of the form f_1 g_1 0 ---> A_1 ----> B_1 ----> C_1 ---> 0 | | | | | | u| v| |w (*) | | | | | | v v v 0 ---> A_0 ----> B_0 ----> C_0 ---> 0 f_0 g_0 The snake lemma says that if both rows are exact, then there is a map from ker w ---> coker u such that 0 -> ker u --> ker v --> ker w --> coker u --> coker v --> coker w -> 0 is exact. What might be called a hidden snake comes when the rows of (*) are not exact, but the conclusion is still valid. Here is an example (it appears as an exercise both in Mac Lane's CWM and in my Acyclic Models). Let f: A --> B and g: B --> C be arbitrary morphisms in an abelian category. Then the snake lemma holds for the diagram 1 f 0 ----> A ------> A ------> B ---> 0 | | | | | | f| gf| |g | | | (**) | | | v v v 0 ----> B ------> C ------> C ---> 0 g 1 whose rows are hardly ever exact (only if f and g are 0, in fact). If the snake sequence, even without the connecting homomorphism, it must be that the pair (g_1f_1, g_0f_0), considered as map of chain complexes of length 2, is null homologous. I have not seen any way of just using that hypothesis, but if we suppose that it is null homotopic, then there is an approach. To be null homotopic requires the existence of a map t: A_0 --> B_1 such that wt = g_0f_0 and tu = g_1f_1. In the case of (*) (or any case where both g_1f_1 and g_0f_0 are 0), you can take t = 0. In the case of (**), you can take t = identity on B. Once t exists, it is possible to write a sequence of length 4 whose exactness implies the exactness of the snake. I do not know if that is a necessary condition, but both (*) and (**) satisfy that condition.