The motivation of this question can be found in
Can we extend the map $φ$ to $ℝ^{r}×C(ℚ)^{\text{tors}}→C(ℚ)$ as an isomorphism or not?
My question is: How I can find the inverse of the isomorphism given in the above link?
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Actually you want to prove the following: if $G$ is an abelian group and $H,K$ are subgroups of $G$ with $G=H+K$ and $H\cap K={0}$, then $G\simeq H\times K$. Isn't this obvious? – Apr 04 '13 at 14:52
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Yes, it obvious. But I want to see the epression of the inverse map of the isomorphism given in the link. – Safwane Apr 04 '13 at 14:54
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Then try to define the group homomorphisms $G\to H\times K$ and $H\times K\to G$ in this general frame. In your particular case this will give you exactly what you want. – Apr 04 '13 at 14:57
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I see only the second one. – Safwane Apr 04 '13 at 14:58
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If $G$ is an abelian group and $H,K$ are subgroups of $G$ such that $G=H+K$ and $H\cap K=\{0\},$ then there is an isomorphism $\phi:G\to H\times K$.
The way to define $\phi$ is the following: for $x\in G$ there exist $h\in H$ and $k\in K$ such that $x=h+k.$ Furthermore, $h,k$ are unique with this property: if $x=h'+k'$, then $h+k=h'+k'$ and thus $h-h'=k'-k\in H\cap K$, therefore $h=h'$ and $k=k'$. Now define $\phi(x)=(h,k)$. (In your particular case $\phi(\sum_{i=1}^r\alpha_iP_i+T)=(\alpha_1,\dots,\alpha_r,T)$.)