$0 \rightarrow{}2\mathbb{Z} \xrightarrow{f} \mathbb{Z} \xrightarrow{p} \mathbb{Z}/\mathbb{2Z} \to 0$ is an exact sequence .True/False

Question

Is the following statement True/False

The sequence $0 \rightarrow{}2\mathbb{Z} \xrightarrow{f} \mathbb{Z} \xrightarrow{p} \mathbb{Z}/\mathbb{2Z} \to 0$ is an exact sequence where $f$ is the inclusion map and $p$ is the projection of $\mathbb{Z} $ onto $ \mathbb{Z}/2\mathbb{Z}$

My attempt : I think this statement is false

Here $Ker p=\{ a \in \mathbb{Z} |p(a)=2\mathbb{Z} \}=\{a \in \mathbb{Z} |a+2\mathbb{Z}=2\mathbb{Z}\}=\{ a\in \mathbb{Z}|a \in 2\mathbb{Z}\}=2\mathbb{Z}$

But $f:2\mathbb{Z} \to \mathbb{Z}$ i,e image of $f$ is $\mathbb{Z}$

Therefore $Im f\neq ker p \implies$ sequence $0 \rightarrow{}2\mathbb{Z} \xrightarrow{f} \mathbb{Z} \xrightarrow{p} \mathbb{Z}/\mathbb{2Z}$ is not an exact sequence

How is the inclusion map from $2 \mathbb{Z}$ to $\mathbb{Z}$ a surjection? — Matthew Leingang, Sep 28 '22 at 15:53
Well, that's not the inclusion map then, that's some other map — Matthew Leingang, Sep 28 '22 at 15:58
okay @MatthewLeingang that means $f(z)=z$ is the inclusion map .Am i right ? — wasiu, Sep 28 '22 at 16:02
$f:2\Bbb Z \to \Bbb Z$ means that the codomain of $f$ is $\Bbb Z$. That's not the image. — jjagmath, Sep 28 '22 at 16:05
This question shouldn't have downvotes; it's an honest question arising from a misconception about the inclusion map. — Matthew Leingang, Sep 30 '22 at 13:14

score 1 · Accepted Answer · answered Sep 28 '22 at 16:58

The kernel of $f$ is $0$, which is the image of the map $0 \to 2\mathbb{Z}$. The kernel of $p$ is $2\mathbb{Z}$, which is the image of $f$. The kernel of $\mathbb{Z}/2\mathbb{Z} \to 0$ is $\mathbb{Z}/2\mathbb{Z}$, which is the image of $p$. So this sequence is exact.

(However, this exact sequence is an example of one that is not split exact.)

$0 \rightarrow{}2\mathbb{Z} \xrightarrow{f} \mathbb{Z} \xrightarrow{p} \mathbb{Z}/\mathbb{2Z} \to 0$ is an exact sequence .True/False

1 Answers1