$0\rightarrow \mathbb{Z} \xrightarrow{ f_k} \mathbb{Z}\ \xrightarrow{ \pi } \mathbb{Z}/{k \mathbb{Z} } \rightarrow 0.$ is exact but not split

Question

let $f_k : \mathbb{Z} \rightarrow \mathbb{Z} $ and $k\in\mathbb{N}$ ,$f_k(x)=kx,\forall x\in\mathbb{Z} $ be homomorphism. now show that

$0\rightarrow \mathbb{Z} \xrightarrow{ f_k} \mathbb{Z}\ \xrightarrow{ \pi } \mathbb{Z}/{k \mathbb{Z} } \rightarrow 0$ is exact but not split. because $f_k$ is one to one and $\pi$ is Surjective function then this sequence is exact.We know this can't split because the Splitting Lemma would then imply $\mathbb{Z} \cong \mathbb{Z} \oplus \mathbb{Z}_k$, which is a contradiction because $\mathbb{Z}$ has no torsion.

Answer 1

In order to show that the sequence is exact you actually need to show that image of each morphism is the kernel of the consecutive one. Since the first morphism is zero then the condition at first element is equivalent to "$f_k$ is injective". And since the last one is zero then the condition is equivalent to "$\pi$ is surjective". But you've missed the middle term, you also need to show that $ker(\pi)=im(f_k)$.

Note however that there's a special case of $k=0$. In this situation $f_0$ is not injective (it is the zero morphism). And so the sequence is not exact.

As for splitting (note that it only makes sense for $k\neq 0$). Yes, you can apply the Splitting Lemma. You can also approach this directly: how many homomorphisms $\mathbb{Z}/k\mathbb{Z}\to\mathbb{Z}$ are there?

However you do have to consider two special cases as well, namely $k=\pm1$. Because in this situation we get that $f_{\pm 1}$ is an isomorphism, $\mathbb{Z}/k\mathbb{Z}=0$, $\pi$ is the zero morphism and the sequence splits.

Answer 2

You don't really need to use the Splitting Lemma here. If the sequence splits then, by definition, there exists a homomorphism $\psi:\mathbb Z\to\mathbb Z$ such that $\psi\circ f_k=\mathrm{Id}_{\mathbb Z}$. But this means that, in particular, that $$\psi(k)=\psi\circ f_k(1) =\mathrm{Id}_{\mathbb Z}(1)=1.$$ But $$\psi(k)=\underbrace{\psi(1)+\cdots+\psi(1)}_{k\text{ times}}=k\psi(1)=1.$$ This is only possible if $k,\psi(1)\in\lbrace\pm 1\rbrace$.