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Answers for “如何从正合序列证明 $\operatorname{coker}(i)\simeq Y$?”

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The proof uses the definition of exactness and the First Isomorphism Theorem. From exactness at ZZ, im⁡(i)=ker⁡(p)\operatorname{im}(i) = \ker(p).

Conditions: X→iZ→pY→0X \xrightarrow{i} Z \xrightarrow{p} Y \to 0 is an exact sequence of RR-modules.; RR is a commutative ring.