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The proposition 2.9 of Atiyah and Macdonald :

It is written that

The sequence $$M'\xrightarrow u M \xrightarrow v M'' \rightarrow 0$$ is exact iff for all $A$-modules N, the sequence $$0\rightarrow Hom (M'',N)\xrightarrow{\bar{v}} Hom(M,N)\xrightarrow{\bar{u}} Hom (M',N)$$ is exact

Proof : First of all ,since $\bar v$ is injective for all $N$ it follows that $v$ is surjective .Next , we have $\bar u \circ \bar v =0$ that is $v \circ u \circ f =0$ for all $f : M'' \to N$

My confusion : Im not getting why is $\bar u \circ \bar v =v \circ u \circ f?$

My thinking :$\bar u \circ \bar v=\bar u (\bar v )$

$$\bar v: Hom (M'',N)\rightarrow Hom(M,N)$$

$$\bar u:Hom(M,N)\rightarrow Hom (M',N)$$

$$\bar u \circ \bar v:Hom (M'',N) \to Hom (M',N)$$

$$\implies \bar u \circ \bar v \in Hom (M',N) \implies \bar u \circ \bar v: M' \to N $$

I think $\bar u \circ \bar v \neq v \circ u \circ f$

wasiu
  • 421

1 Answers1

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There is a typo, it should be

Next, we have $\bar u \circ \bar v = 0$, that is $f \circ v \circ u = 0$ for all $f \colon M’’ \to N$.

Why? Well, if $\bar u \circ \bar v \colon \text{Hom}(M’’,N) \to \text{Hom}(M’,N)$ is the zero map, then for every $f \in \text{Hom}(M’’,N)$ we have that $$ 0 = (\bar u \circ \bar v)(f) = \bar u(\bar v(f)) = \bar u(f \circ v) = f \circ v \circ u. $$

azif00
  • 20,792