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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.

share|improve this question
    
$\overline{d_1}$ doesn't have to be injective for the sequence to be exact at $\textrm{Hom}(P_1, N)$. However, you are right that the sequence need not be exact at $\textrm{Hom}(P_1, N)$. (If exactness was preserved in the way you suggest, then $\textrm{Hom}$ would be an exact functor!) –  Zhen Lin Jul 22 '12 at 8:23
    
@ZhenLin I think you are misreading, I am not suggesting exactness of $\mathrm{Hom}$, I am requesting a concrete example. –  Rudy the Reindeer Jul 22 '12 at 8:33

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
For the truncated sequence I get $$ 0 \to 2 \mathbb Z \hookrightarrow \mathbb Z \to 0$$ and applying $\mathrm{Hom}$ yields $$ 0 \to \mathrm{Hom}( \mathbb Z , \mathbb Z / 2 \mathbb Z) \to \mathrm{Hom} (2 \mathbb Z , \mathbb Z / 2 \mathbb Z ) \to 0$$ where I think $\mathrm{Hom} (2 \mathbb Z , \mathbb Z / 2 \mathbb Z ) = 0$ but you say $\mathrm{Hom} (2 \mathbb Z , \mathbb Z / 2 \mathbb Z ) = \mathbb Z / 2 \mathbb Z$. Is that a typo? –  Rudy the Reindeer Jul 22 '12 at 10:49
1  
$2 \mathbb{Z} \cong \mathbb{Z}$, so we must have $\textrm{Hom}(2 \mathbb{Z}, \mathbb{Z} / 2 \mathbb{Z}) \cong \textrm{Hom}(\mathbb{Z}, \mathbb{Z} / 2 \mathbb{Z})$. –  Zhen Lin Jul 22 '12 at 11:18
1  
@MattN. Any group homomorphism $\phi \in \textrm{Hom}(2\Bbb{Z}, \Bbb{Z}/2\Bbb{Z})$ must send $0$ to $0$. Now if you send $2$ to $0$ by additivity $\phi$ is the zero map, and if you send $2$ to $1$ again by additivity everything else is sent to is completely determined. So you actually have two choices and not one and so it cannot be the trivial group. –  user38268 Jul 22 '12 at 14:40
    
@BenjaLim Thanks Ben, this comment is useful. So we have $\mathrm{Hom}(n \mathbb Z, N) \cong \mathrm{Hom}(\mathbb Z, N)$ for a $\mathbb Z$-module $N$? –  Rudy the Reindeer Jul 28 '12 at 19:11

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