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 compactβopen 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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3$\begingroup$ Is there a reason you take this $U$ as your domain? It would seem more natural to work with the unit disk, say. $\endgroup$– Christian RemlingCommented May 26, 2019 at 20:27
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$\begingroup$ +1 to Christian Remling. You will then have $\{f: D\to F, f(0)=Q\}$ $\endgroup$– erzCommented May 26, 2019 at 23:56
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$\begingroup$ If $F={\mathbb C}$ then your space is homotopy equivalent to $S^1$. Most likely, this is always the case. $\endgroup$– Moishe KohanCommented May 27, 2019 at 1:49
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1$\begingroup$ In the case when πΉ=β (assuming the normalization at π=0 and π§β unit disk) one uses the isotopy given by π(π§,π‘)=π(π‘π§)/π‘, π‘β(0,1]. As π‘β0, this converges to the linear function π§β¦πβ²(0)π§. Thus, π 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. $\endgroup$– Moishe KohanCommented May 27, 2019 at 4:56
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$\begingroup$ Oh okay β¦ 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? $\endgroup$– FKranholdCommented May 27, 2019 at 5:47
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