# how the Snake Lemma has been applied here?

In the proof of the following Lemma of The Stacks Project, it is mentioned that they used Snake Lemma as follows:

However, I don't see how the Snake Lemma has been applied here ?

Because the Snake Lemma gives the following conclusion

$$(1)$$ the sequence $$\ker(f) \xrightarrow{a} \ker(g) \xrightarrow{b} \ker(h) \longrightarrow \operatorname{coker}(f) \xrightarrow{c} \operatorname{coker}(g) \xrightarrow{d} \operatorname{coker}(h)$$ is exact,

$$(2)$$ if $$F(A) \to F(A) \oplus F(B)$$ injective then $$a$$ is injective and if $$F(A \oplus B) \to F(B)$$ is surjective then $$d$$ is surjective.

But how does these two conclusions helps here?

Note that in this case $$f=\operatorname{id}_{F(A)}$$ and $$h=\operatorname{id}_{F(B)}$$. Hence

$$\ker f=\ker h=0,\ \operatorname{coker}f=\operatorname{coker}h=0$$

This gives exact sequences

$$\ker f=0\rightarrow\ker g\rightarrow 0=\ker h,\ \operatorname{coker} f=0\rightarrow\operatorname{coker} g\rightarrow 0=\operatorname{coker} h$$

But its a standard fact that $$0\rightarrow M\rightarrow 0$$ being exact implies that $$M=0$$. Hence $$g$$ is an isomorphism as well since we have that $$\ker g=\operatorname{coker}g=0$$. So the full strength of the snake lemma is not needed here (as these kernel and cokernel sequences are rather trivial to construct) but rather an important byproduct.

• Ok, so, $0 \to ker(g) \to 0$ exact implies ker$(g)=0$ implies $g$ is injective while $0 \to coker(g) \to 0$ exact implies $g$ is surjective and hence $g$ is bijective.
– MAS
Commented Mar 28, 2021 at 13:39
• @Masmath Exactly. Commented Mar 28, 2021 at 13:40
• Thank you very much
– MAS
Commented Mar 28, 2021 at 13:41