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$D(P,r)\backslash \{P\}\xrightarrow{f}U\xrightarrow{g}\mathbb{C}$. Both $f$ and $g$ are holomorphic. If $f$ has a removable (or pole, or essential) singularity at $P$, does $g\circ f$ also have one?

I think in any case $g\circ f$ will not have singularities any more. It is just a hunch, and I cannot explain it clearly.

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    Thank Lukas and Old John for your hints and details. If $f$ has essential singularity at $P$, by Casorati-Weierstrass theorem, $f(D(P,r)\backslash \{P\})=U$ is dense in $\mathbb{C}$. Does it mean $g$ has no singularity is this case?2012-10-31

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A couple of things to consider:

If $g$ is the function $g:z \mapsto z$, then the composite will have exactly the same behaviour at $P$ as $f$ does.

At the other extreme, if $g$ is the constant function that maps everything to zero, then the composite function will have at worst a removable singularity, depending on whether you consider $\infty$ to be a point in the co-domain of $g$.