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Let $F$ be a Riemann surface and $Q\in F$. Consider $U:=(\mathbb{C}\cup\infty)\setminus [-1,1]^2$. I am interested in the space $$X:=\{f:U\to F;\,\text{$f:U\to f(U)$ biholomorphic and $f(\infty)=Q$}\},$$ endowed with the compact–open topology. I would expect $X$ to be contractible because $U$ is contractible, but I simply know too little about these kind of spaces.

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    $\begingroup$ Is there a reason you take this $U$ as your domain? It would seem more natural to work with the unit disk, say. $\endgroup$ Commented May 26, 2019 at 20:27
  • $\begingroup$ +1 to Christian Remling. You will then have $\{f: D\to F, f(0)=Q\}$ $\endgroup$
    – erz
    Commented May 26, 2019 at 23:56
  • $\begingroup$ If $F={\mathbb C}$ then your space is homotopy equivalent to $S^1$. Most likely, this is always the case. $\endgroup$ Commented May 27, 2019 at 1:49
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    $\begingroup$ In the case when 𝐹=β„‚ (assuming the normalization at 𝑄=0 and π‘§βˆˆ unit disk) one uses the isotopy given by 𝑓(𝑧,𝑑)=𝑓(𝑑𝑧)/𝑑, π‘‘βˆˆ(0,1]. As 𝑑→0, this converges to the linear function 𝑧↦𝑓′(0)𝑧. Thus, 𝑋 deformation retracts to ${\mathbb C}^\times$ (so it is not contractible). The same argument works when $F$ is the unit disk. Some version of this argument might also work for general surfaces. $\endgroup$ Commented May 27, 2019 at 4:56
  • $\begingroup$ Oh okay … so if I additionally fix a tangent vector $0\ne v\in T_QF$ and demand that $f'(0)=v$, then $X$ would be contractible? $\endgroup$
    – FKranhold
    Commented May 27, 2019 at 5:47

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