My question comes from reading Munkres' Topology, the section on Stone-Čech compactification.
To find the compactification $\mathrm{Y}$ of $\mathrm{X}$, we find an embedding h,
$\mathrm{h}: X \rightarrow Z$
such that $\mathrm{Z}$ is compact Hausdorff.
$\overline{h(X)}$ is compact in $\mathrm{Z}$. To construct $\mathrm{Y}$, we look at the points added by taking this closure. Then we let $\mathrm{Y}=X \cup A$, where $\mathrm{A}$ is disjoint from X and in bijection with $\overline{h(X)}\setminus{h(X)}$. We give Y a topology by considering the bijection:
$\mathrm{H:Y\rightarrow Z}$ given by $\mathrm{H\rvert}_X\mathrm{=h}$ and $\mathrm{H\rvert}_A$ to be the bijection of $A$ with $\overline{h(X)}\setminus{h(X)}$
and we say $U \subset Y$ is open if $H(U)$ is open. So H is an embedding.
My confusion comes from when I try to extend a continuous real-valued function on $X$ to $Y$. I'm not sure what to do with the points I added, namely those in the set $A$ as defined above.
Here is an example of my thought process.
Let $f: (0,1) \rightarrow \Bbb{R}, f(x)={x}^{2}-x$.
The compactification of X will be the space Y obtained via its 1-point compactification, so I can imagine an embedding to ${S}^{1}$, and the closure just "adds" the single missing point. Intuitively, I guess that I can extend $f$ to $\tilde{f}$ iff the limits of $f$ at 0 and 1 agree. This is because I'm thinking of (0,1) as its image in $S^1$, and I know that if I want $\tilde{f}$ to be continuous, I only have hope if their limits agree, since I've "glued together" 0 and 1. So I can extend f by defining $\tilde{f}(a)=0$, where a is the point picked up from the 1 point compactification.
My problem is that I don't know how to do this for more complicated embeddings. Instead of embedding into the circle, let's use the map $h: (0,1) \rightarrow \Bbb{R}^{2}$ by $h(x)=(x,\sin(1/x))$. The closure in $\Bbb{R}^{2}$ adds the points $(1,\sin(1))$ and the region $\{(0,y): -1 \leq y \leq 1\}$.
How can I determine conditions (besides boundedness) of $f$ such that I can extend it continuously? I'm not sure how to think of the compactification (0,1) since the map isn't as simple as gluing 0 to 1.
My exercise was to show $\cos(1/x)$ could not be continuously extended, and I realized I don't know the right way to think about the "compactified" space.