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