Is $operatorname{coker}colonmathcal C^{[1]}tomathcal C$ a faithful functor?
Multi tool use
$begingroup$
Let $mathcal C$ be an abelian category; e.g., $mathcal C=operatorname{mathit{A}-Mod}$. Let $mathcal C^{[1]}$ denote the category of morphisms in $mathcal C$, where morphisms $mathcal C^{[1]}$ are commutative squares in $mathcal C$.
Taking cokernels yields a functor $operatorname{coker}colonmathcal C^{[1]}tomathcal C$. In the example that $mathcal C$ is the category of $A$-modules, we know that we can find a free presentation for every $A$-module, which is an object in $mathcal C^{[1]}$. Every morphism of $A$-modules can be lifted to a morphism of the free presentations. Hence, $operatorname{coker}$ is a full functor when restricted to $(text{free $A$-modules})^{[1]}$.
Question: Is this functor also faithful?
commutative-algebra
asked Jan 10 at 13:29
BubayaBubaya
398111
$endgroup$
add a comment |
$begingroup$
Let $mathcal C$ be an abelian category; e.g., $mathcal C=operatorname{mathit{A}-Mod}$. Let $mathcal C^{[1]}$ denote the category of morphisms in $mathcal C$, where morphisms $mathcal C^{[1]}$ are commutative squares in $mathcal C$.
Taking cokernels yields a functor $operatorname{coker}colonmathcal C^{[1]}tomathcal C$. In the example that $mathcal C$ is the category of $A$-modules, we know that we can find a free presentation for every $A$-module, which is an object in $mathcal C^{[1]}$. Every morphism of $A$-modules can be lifted to a morphism of the free presentations. Hence, $operatorname{coker}$ is a full functor when restricted to $(text{free $A$-modules})^{[1]}$.
Question: Is this functor also faithful?
commutative-algebra
asked Jan 10 at 13:29
BubayaBubaya
398111
$endgroup$
add a comment |
$begingroup$
Let $mathcal C$ be an abelian category; e.g., $mathcal C=operatorname{mathit{A}-Mod}$. Let $mathcal C^{[1]}$ denote the category of morphisms in $mathcal C$, where morphisms $mathcal C^{[1]}$ are commutative squares in $mathcal C$.
Taking cokernels yields a functor $operatorname{coker}colonmathcal C^{[1]}tomathcal C$. In the example that $mathcal C$ is the category of $A$-modules, we know that we can find a free presentation for every $A$-module, which is an object in $mathcal C^{[1]}$. Every morphism of $A$-modules can be lifted to a morphism of the free presentations. Hence, $operatorname{coker}$ is a full functor when restricted to $(text{free $A$-modules})^{[1]}$.
Question: Is this functor also faithful?
commutative-algebra
asked Jan 10 at 13:29
BubayaBubaya
398111
$endgroup$
Let $mathcal C$ be an abelian category; e.g., $mathcal C=operatorname{mathit{A}-Mod}$. Let $mathcal C^{[1]}$ denote the category of morphisms in $mathcal C$, where morphisms $mathcal C^{[1]}$ are commutative squares in $mathcal C$.
Taking cokernels yields a functor $operatorname{coker}colonmathcal C^{[1]}tomathcal C$. In the example that $mathcal C$ is the category of $A$-modules, we know that we can find a free presentation for every $A$-module, which is an object in $mathcal C^{[1]}$. Every morphism of $A$-modules can be lifted to a morphism of the free presentations. Hence, $operatorname{coker}$ is a full functor when restricted to $(text{free $A$-modules})^{[1]}$.
Question: Is this functor also faithful?
commutative-algebra
commutative-algebra
asked Jan 10 at 13:29
BubayaBubaya
398111
asked Jan 10 at 13:29
BubayaBubaya
398111
asked Jan 10 at 13:29
BubayaBubaya
398111
asked Jan 10 at 13:29
BubayaBubaya
398111
asked Jan 10 at 13:29
BubayaBubaya
398111
398111
add a comment |
add a comment |
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