If an immersion $X$ maps circles into planes then its image $X(\mathbb{D})$ is homeomorphic to the cylinder.

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Let $X:\left( u,v\right)\in \mathbb{D}\backslash \left\{ 0\right\}\subseteq\mathbb{R}^2 \mathbb{% \longmapsto }\left( x\left( u,v\right) ,y\left( u,v\right) ,z\left( u,v\right) \right) \in \mathbb{\mathbb{R}}^{3}$ an minimal immersion, where $\mathbb{D=}\left\{ p\in \mathbb{R}^{2};\left\Vert p\right\Vert <1\right\} $ is unitary open disc.

If $X$ maps circles into planes (coordinate function $z$ constant) then $X(\mathbb{D})$ is homeomorphic to cylindric.

I do not know if all these assumptions are necessary, but that's what I have.

Can you help me prove that? I studied geometry and topology a long time ago.