Pick an element "x" in ker "c" and view it as an element of "C"; since "g" is surjective, there exists "y" in "B" with "g"("y") = "x". Because of the commutativity of the diagram, we have "g"'("b"("y")) = "c"("g"("y")) = "c"("x") = 0 (since "x" is in the kernel of "c"), and therefore "b"("y") is in the kernel of "g' ". Since the bottom row is exact, we find an element "z" in "A' " with "f" '("z") = "b"("y"). "z" is unique by injectivity of "f" '. We then define "d"("x") = "z" + "im"("a"). Now one has to check that "d" is well-defined (i.e., "d"("x") only depends on "x" and not on the choice of "y"), that it is a homomorphism, and that the resulting long sequence is indeed exact. One may routinely verify the exactness by diagram chasing (see the proof of Lemma 9.1 in ).