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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 $\partial f=a'-a$?

Hang
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  • When $a'-a$ is the unit circle traversed twice and $\sigma_n = { |z|< 1, arg(z)\in (2n2\pi/3,(2n+2)2\pi/3)},n\in 0,1,2$ what gluing do you expect ? – reuns Sep 28 '19 at 23:39
  • @reuns Let's say we can make choices of $b$ and $\sigma_i$, they are not given. The condition is only that $a'-a$ is homologous to zero. Thanks for the clarification, I've edited my post. – Hang Sep 29 '19 at 01:02
  • It depends on what you mean by a "surface". If you mean an embedded surface then of course not. – Moishe Kohan Sep 29 '19 at 04:50
  • @MoisheKohan I think I just need a continuous map. Roughly, I want to glue the domains $\Delta^2$ of 2-simplices to get a surface $S=\sqcup_k \Delta^2$. – Hang Sep 29 '19 at 14:13
  • @Hang Then it exists for any topological space. Hatcher explains it nicely in his book. – Moishe Kohan Sep 29 '19 at 22:55

1 Answers1

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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.)

Moishe Kohan
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