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In geometric topology, one often encounters the following argument(or alike):

If two (singular) 1-cycles $a_1,a_2:\Delta \to X$ are homologous, then there exists a bordism(singular surface) $F$ that joins $a_1$ and $a_2$. Thus, if one wants to prove that a homomorphism from the singular chain complex to an abelian group can quotient out to give an induced homomorphism from the first homology group, it suffices to show the homomorphism is invariant under any cobordism.

(My apology if there's a mistake in the statement above, but I think the spirit of the argument is more or less contained.)

E.g. When defining a Seifert form associated to the Seifert surface $\Sigma_g$ as a bilinear form on the 1st homology group of $\Sigma_g$, in lemma 2.6 in the following paper that introduces an isomorphism from $H_1(\mathrm{Sym}^g\Sigma_g)$ to $H_1(\Sigma_g)$, or many other places that I can't refer just now.

As I haven't seen (the proof of) the above argument written explicitly, it might not hold as itself given above, but can anyone suggest a reference or give a proof of this argument? Explicitly, I want to know one of the explicit form of the above argument, made explicit in the range of possible space $X$.(I guess $X$ being manifold should work?)

And also I guess this argument has something to do with the canonical map $\Omega_1^{or}(X)\to H_1(X)$, where $\Omega_1^{ori}(X)$ is the oriented bordism group of $X$. But I'm not familiar with the language in cobordism theory, so I appreciate if this relation can be explained.

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    The bordism can be chosen to be nonsingular. The argument is more or less given in Hatcher's chapter 2 appendix. – 2017-01-12
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    @MikeMiller Then I guess what you mean is the part about Hurewicz map, right? I was once not very convinced why the resulting delta complex is orientable surface, but it is indeed mentioned in Hatcher's(maybe I just missed that line in the first reading), so I'd better try to convince myself. Thanks. – 2017-01-12
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    The resulting Delta complex is not an oriented surface always - for instance, take the cone on two circles. But it can always be modified to be a surface. – 2017-01-12
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    @MikeMiller I'm confused by your example; In the construction of the delta complex $K$, the edges of 2-simplices are glued by pairs, thus if there's some families of edges that has common vertices and whose edges are glued to other edges in the family in which the edge belongs, then your cone-type identification does not happen(in my head). – 2017-01-13

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