Why is $\operatorname{Ext}^0(A,B)\simeq \operatorname{Hom}(A,B)$?

Asked Jul 08 '21 at 05:43

Active Jul 08 '21 at 05:54

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I want to prove

$\operatorname{Ext}^0(A,B)\simeq \operatorname{Hom}(A,B)$ where $A,B$ are modules over some ring $R$.

My definition of $\operatorname{Ext}^n(A,B)=H^n(\operatorname{Hom}(C,B))$ where $\operatorname{Hom}(C,B) = \{\operatorname{Hom}(C_n,B),\delta^n\}$ where $C_n$ is a canonical free resolution of $A$. So I need to show $\operatorname{Hom}(A,B) = \operatorname{ker}(\delta^0)$ for $\operatorname{Hom}(C_0,B)\xrightarrow{\delta^0}\operatorname{Hom}(C_1,B)$. But I can't see where the equivalence rises. Could you help? It seems this fact is accepted as an obvious fact but not to me.

edited Jul 08 '21 at 05:50

HallaSurvivor

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asked Jul 08 '21 at 05:43

one potato two potato

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Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see the guidelines for good use of $\rm\LaTeX$ in question titles. – soupless Jul 08 '21 at 05:49
@Christoph I don't think so. My definition uses free resolution so I want to prove without injective resolution or projective resolution (and book didn't introduced that term.). – one potato two potato Jul 08 '21 at 05:57
@PeterJohn Free resolutions are projective resolutions, so those answers that use projective resolutions should work for you. – Christoph Jul 08 '21 at 06:22
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This question has actually nothing to do with free or projective resolutions. The question simply is: “Why is $\operatorname{Hom}(-,B)$ left exact?” – Claudius Jul 08 '21 at 06:29

Why is $\operatorname{Ext}^0(A,B)\simeq \operatorname{Hom}(A,B)$?

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