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I would like to ask what the smallest Hausdorff compactification of $mathbb{R}$ is, onto which we may continuously extend the function $g: mathbb{R} rightarrow mathbb{R}, x mapsto text{sin}(x)$.

A weaker / related version of this question has been asked here, and even that only has a partial answer: Smallest compactification for continuous extension of $sin(x)$

There the answer is about minimal compactifications; I would like to know about a smallest one. I believe, mimicking the answer given there, that the compactification I am looking for is $h=(f,g): mathbb{R} rightarrow [-1,1] times [-1,1], x mapsto (text{tanh}(x),text{sin}(x)) $.
(The first component is just an embedding of $mathbb{R}$ into an interval.) Let's call this compactification $(h,tilde{mathbb{R}})$. It is my candidate for being a smallest compactification onto which we may extend $g(x)=$ sin$(x)$.

SUMMARY: I would like to show that given any compactification $(j, gamma mathbb{R})$ such that the function $g(x)=$ sin$(x)$ extends continuously onto it, there exists a continuous map $k$ from $gamma mathbb{R}$ to $tilde{mathbb{R}}$ such that $kj=h$.

Idea for proof: I'm sure it suffices (and is necessary!) to have that $f(x)= $ tanh$(x)$ extends continuously to $(j, gamma mathbb{R})$, since by assumption $g(x)$ already extends continuously to it. However the idea I had for constructing an extension of $f(x)$ involves using sequential compactness rather than compactness, which is not allowed because in general the two notions don't coincide. For instance, I've tried to define the extension $tilde{f}: gamma mathbb{R} rightarrow [-1,1]$ as being equal to $f$ on the image of $j$ (obviously!), but then the remaining points of $gamma mathbb{R}$ have to go to either $-1$ or $1$ depending on some condition that I've not worked out. I tried to say, let $tilde{f}$ take the value $1$ (respectively $-1$) on $p in gamma mathbb{R} backslash j(mathbb{R})$ if there exists a sequence of distinct real numbers $(x_n)$ such that cos$(x_n)$ tends to $1$ (respectively $-1$) and $j(x_n)$ tends to $p$, but that's not well-defined for every $p$.

Consider the injective map $jcolonBbb Rto Bbb R^3$, $tmapsto (sin t,cos t, sin sqrt 2t)$ and let $gamma Bbb Rsubset Bbb R^2$ be the closure of the image of $j$. Clearly, $gamma Bbb R$ is the cylinder $S^1times [-1,1]$. Then $gamma Bbb Rto Bbb R$, $(x,y,z)mapsto x$ is a continuous extension of the sine function, but we cannot map the cylinder suitably to $tilde{Bbb R}$ because the points $j(2npi)$ and $j(-2npi)$ are both dense in ${(0,1)}times [-1,1]$