Is there an infinite Hausdorff space $(X,\tau)$ with the following property? 



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> If $x\neq y \in X$ and $f:\{x,y\}\to X$ is a map, then there is exactly one continuous function $f': X\to X$ such that $f'|\_{\{x,y\}} = f$.
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$2$-determined Hausdorff spaces
Using the technique of [van Mill](http://www.ams.org/journals/tran/1983-280-02/S0002-9947-1983-0716833-2/S0002-9947-1983-0716833-2.pdf) it is possible to prove the following 



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> **Theorem.** There exists a subset $Z$ in the complex plane $\mathbb C$ such that 
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> 1) for any complex numbers $x,a,b,c,d\in Z$ with $a\ne b$ and $c\ne d$ we have $h\_{a,b,c,d}(Z)\subset Z$ where $h\_{a,b,c,d}(z):=c+\frac{d-c}{b-a}(z-a)$;
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> 2) for any homeomorphism $h:Z\to Z$ there exist unique complex numbers $a,b,c,d\in Z$ with $a\ne b$ and $c\ne d$ such that $h=h\_{a,b,c,d}$;
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> 3) Under CH (more generally, under $\acute{\mathfrak n}=\mathfrak c$) for any non-constant continuous map $f:Z\to Z$ there exist unique points $a,b,c,d\in Z$ with $a\ne b$ and $c\ne d$ such that $f=h\_{a,b,c,d}$.
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Here by $\acute{\mathfrak n}$ denotes the smallest cardinality of an infinite cover of $[0,1]$ by pairwise disjoint closed sets. By the classical Sierpi$\acute{\mathfrak n}$ski Theorem, $\aleph\_1\le \acute{\mathfrak n}\le\mathfrak c$. More information on the small uncountable cardinal $\acute{\mathfrak n}$ can be found in [this MO-post](https://mathoverflow.net/questions/285780/are-the-sierpi%C5%84ski-cardinal-acute-mathfrak-n-and-its-measure-modification).



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> **Remark.** Under $\acute{\mathfrak n}=\mathfrak c$, the space $Z$ from Theorem has the property required in the question of Dominic van der Zypen: for any distinct points $a,b\in Z$ and point $c,d\in Z$ the map $h\_{a,b,c,d}$ is a unique continuous self-map $f:Z\to Z$ such that $f(a)=c$ and $f(b)=d$.
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