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

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

Because the Snake Lemma gives the following conclusion

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$(1)$ 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?

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1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – MAS
    Commented Mar 28, 2021 at 13:39
  • $\begingroup$ @Masmath Exactly. $\endgroup$
    – mrtaurho
    Commented Mar 28, 2021 at 13:40
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    $\begingroup$ Thank you very much $\endgroup$
    – MAS
    Commented Mar 28, 2021 at 13:41

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