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- Thread starter Mr Davis 97
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fresh_42

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I see only additive groups here, so it has to be ##g+\operatorname{ker}(\pi)##. Moreover we have an epimorphism, so we get an induced isomorphism ##\bar{\pi}\, : \,\mathbb{R} \cong \mathbb{R}^2/\operatorname{ker}(\pi)## which means, all elements are of the form ##g+\mathbb{R}## where ##\mathbb{R}=\operatorname{ker}(\pi)## is the diagonal, which is shifted upwards (##g>0##) or downwards (##g<0##) to cover the entire plane by copies of said diagonal.

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