Let $X$ be a good topological space and let $a,a': [0,1]\to X$ be two paths in it with the same end points $a(0)=a'(0)=p$ and $a(1)=a'(1)=q$. In particular, $a'-a$ is a cycle.
Now, suppose $a'-a$ is homologous to zero.
Can we always find a surface $S$ and a continuous map $f:S\to X$ with $\partial f=a'-a$?
I will be using singular homology with integer coefficients throughout the answer. See Hatcher's "Algebraic Topology", pages 108-109, for an explanation of the statements below. See also my answer here.
Below, $X$ is an arbitrary topological space.
(a) Suppose that $[c]\in H_1(X)$ is a homology class. Then there exists a continuous map $f: S^1\to X$ such that the image of the fundamental class of $S^1$ under $f$ is $[c]$.
This statement holds even on chain-level: Suppose that $c=\sum_{i=1}^n \sigma_i$, where $\sigma_i: [0,1]\to X$ are singular chains. Assume, in addition, that $c$ is a cycle. Then there exists a map $f: S^1\to X$ and a subdivision of $S^1$ in $n$ (oriented) arcs $\alpha_1,..,\alpha_n$ such that, after composing $f$ with orientation-preserving parameterizations $[0,1]\to \alpha_i$, we have $$ \sigma_i=f|_{\alpha_i}, i=1,...,n. $$
(b) Suppose, in addition, that $[c]=0$ in $H_1(X)$. Then there exists a compact oriented triangulated surface $S$ with one boundary component (identified with $S^1$ above) such that each arc $\alpha_i$ above is an edge of the triangulation and a map $F: S\to X$ extending $f$ as above.
Informally, every null-homologous cycle in $X$ bounds a "surface" in $X$.
The same works for null-homologous maps of surfaces in $X$, and null-homologous maps of 3-manifolds to $X$. However, it does not work for maps of 4-dimensional manifolds (even if $X$ is a single point.)