As you said, functions define relations between the different "dimensions" of the space. This idea is made explicit via correlation functions in Gaussian processes (this is an excellent non-engineering introduction) which can be framed in the theory of Reproducing Hilbert Kernel Spaces (RHKS).
Based on your comment, I think you might want to look at the article,
Schaback, R., & Wendland, H. (2006). Kernel techniques: from machine
learning to meshless methods. Acta Numerica, 1–97.
http://doi.org/10.1017/S0962492904000077
Therein Sobolev spaces are studied in relation with their RHKS. This is an engineering article and not a mathematics one; nevertheless, it provides lots of references for further reading.
A homomorphism can be from a bigger to a smaller graph. Here’s a concrete example:
a
/ \
/ \ 0
b c / \
| | / \
d-----e 1-----2
G H
The map $h:V(G)\to V(H)$ given by $h(a)=h(d)=0$, $h(b)=h(c)=1$, and $h(e)=2$ is a graph homomorphism: it sends edges $ab,db$, and $ac$ to $01$, $ce$ to $12$, and $de$ to $02$. Clearly, however, $G$ and $H$ are not isomorphic, since they don’t even have the same number of vertices.
Notice that in general if $h:V(G)\to V(H)$ is a graph isomorphism, then $h(u)h(v)$ is an edge of $H$ if and only if $uv$ is an edge of $G$. If $h$ is merely a graph homomorphism, however, $h(u)h(v)$ can be an edge of $H$ even if $uv$ isn’t an edge of $G$. Thus, if $K$ is the graph shown below, the map that takes $a,b,c,d$, and $e$ to $0,1,2,3$, and $4$, respectively, is a homomorphism but not an isomorphism.
0
/ \
/ \
1-----2
| |
3-----4
K
Best Answer
If you have a simple path between two points in the disk, you can morph it continuously into any other path in the disk, so homotopically all paths in the disk are equivalent.
However, in an annulus, there are paths that cannot be morphed continuously into each other (while keeping the same endpoints - the hole gets in the way), so they aren't equivalent. Think about paths that wind around the hole vs paths that don't.