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Let $ P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \xrightarrow{d_0} M \to 0$

be an exact sequence of $R$-modules. Consider

$ (*) \hspace{1 cm} P_2 \xrightarrow{d_2} P_1 \xrightarrow{d_1} P_0 \to 0$

that is, the sequence with $M$ removed. Then this resulting sequence should still be exact at $P_1$ since we did not change the maps $d_2, d_1$. Now apply the left exact functor $\mathrm{Hom}(-, N)$ (for some $R$-module $N$) to get

$ (**) \hspace{1 cm} 0 \to \mathrm{Hom}(P_0, N) \xrightarrow{\overline{d_1}} \mathrm{Hom}(P_1, N) \xrightarrow{\overline{d_2}} \mathrm{Hom}(P_2, N) $

Clearly, $\overline{d_1}$ does not have to be injective since $(*)$ was not exact at $P_0$. What I'm not so clear about is why, even though $(*)$ was exact at $P_1$, we also don't necessarily get exactness at $\mathrm{Hom}(P_1, N)$ anymore.

Can you give me a simple example with concrete $R$-modules $P_1, M, N$ such that $(*)$ is exact at $P_1$ but $(**)$ not exact at $\mathrm{Hom}(P_1, N)$? Thanks.

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    @ZhenLin I think you are misreading, I am not suggesting exactness of $\mathrm{Hom}$, I am requesting a concrete example.2012-07-22

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Let's take $R = \mathbb{Z}$, $M = \mathbb{Z} / 2 \mathbb{Z}$, $P_0 = \mathbb{Z}$, $P_1 = 2 \mathbb{Z}$, $P_2 = 0$. There is an evident exact sequence $0 \longrightarrow 2 \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} / 2 \mathbb{Z} \longrightarrow 0$ and applying $\textrm{Hom}(-, \mathbb{Z} / 2 \mathbb{Z})$ to the truncation yields the sequence $\mathbb{Z} / 2 \mathbb{Z} \stackrel{0}{\longrightarrow} \mathbb{Z} / 2 \mathbb{Z}\longrightarrow 0$ which fails to be exact anywhere at all.

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    @RudytheReindeer Yes, functors send isomorphisms to isomorphisms.2017-10-03