morphism

**sfspitfire23** · October 22nd 2009, 07:51 PM

If I have a mapping under addition $\alpha : \mathbb{z} \rightarrow \mathbb{z}$ given by $\alpha(h(x))=h'(x)$

to show it would I do this-

$\alpha(h(x)+k(x))=\alpha h(x) + \alpha k(x)$

Then when mapping is applied $h'(x)+k'(x)=h'(x)+k'(x)$ and this shows that $\alpha : \mathbb{z} \rightarrow \mathbb{z} \alpha(h(x))=h'(x)$ is a homomorphism? or is there more?

**el_manco** · October 22nd 2009, 11:44 PM

Hello

Your proof is correct. I suppose that your are considering $F$ only with que strucuture of group given by the usual adittion of funcitons.

Best regards.

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