I've just been introduced to the Eilenberg-Steenrod axioms, and as an example, we are constructing cellular homology. Let $X$ be a CW complex and define $C_n(X)$ to be the free abelian group generated by $n$-cells of $X$. Define boundary homomorphisms $C_{n}(X)\to C_{n-1}(X)$ by, where $e_\alpha^n$ is an $n$-cell of $X$ and $e^{n-1}_\beta$ an $(n-1)$-cell $$d_n[e^n_\alpha]=\sum_\beta c_{\alpha\beta} [e^{n-1}_\beta]$$ Define the coefficients $c_{\alpha\beta}$ to be the degree of the map $$S^{n-1}\xrightarrow{\chi_\alpha} X_{n-1}\to X_{n-1}/(X_{n-1}\setminus e^{n-1}_\beta)\to S^{n-1}$$ Where $\chi_\alpha$ is the attaching map of $e^n_\alpha$, the second map collapses everything but $e^{n-1}_\beta$ to a point, and the last map is the characteristic map of $e^{n-1}_\beta$. My question is, how does this give a chain complex? I can't easily see that $d^2=0$.


1 Answer 1


Here's a rough outline. You can find this explained precisely in Hatcher's "Algebraic Topology", Section 2.2.

Rather than defining $C_n(X)$ in the informal manner you describe, it is better to define it formally as the relative singular homology group $$C_n(X) = H_n(X_n,X_{n-1}) $$ You can then prove that $C_n(X)$ is a free abelian group, with a basis element for each $n$ cell $e^n_\alpha$, where that basis element can be described explicitly as the image of the generator of $H_n(D^n,S^{n-1}) \approx \mathbb Z$ under the homomorphism induced by the characteristic map $x^n_\alpha : (D^n,S^{n-1}) \to (X_n,X_{n-1})$ of the cell $e^n_\alpha$. And then you can prove that the boundary homomorphism you describe is identical to a connecting homomorphism in the long exact sequence of the triple $(X_n,X_{n-1},X_{n-2})$, namely $$H_n(X_n,X_{n-1}) \mapsto H_{n-1}(X_{n-1},X_{n-2}) $$ And, finally, you can do a diagram chase to prove that the composition of two successive boundary homomorphisms factors through the composition of two successive terms of an exact sequence, namely through $$H_n(X^n) \mapsto H_n(X^n,X^{n-1}) \mapsto H_{n-1}(X^{n-1}) $$ I cannot quite reproduce the diagram here because you can't draw diagonal arrows in mathjax, but here's the best I can do: $\require{AMScd}$ \begin{CD} H_n(X^n) \\ @AAA \\ H_{n+1}(X^{n+1},X^{n}) @>>> H_n(X^n,X^{n-1}) @>>> H_{n-1}(X^{n-1},X^{n-2}) \\ @. @. @AAA \\ @. @. H_{n-1}(X^{n-1}) \end{CD} The vertical arrows are homomorphisms of certain long exact sequences of pairs. If you draw in the two diagonal arrows of negative slope then you'll get two commutative triangles and a 3 term exact sequence of negative slope.

  • $\begingroup$ Thanks for the detailed answer, but we were introduced to homology with the axioms, and haven't done singular homology at all yet. I know this is a (very) nonstandard approach to this but I was wondering if there was another way to see this (the lecturer seemed to think it wasn't difficult). Thank you! $\endgroup$ Apr 2, 2021 at 0:40
  • $\begingroup$ Well, I agree with the lecturer, because the singular homology arguments aren't difficult. $\endgroup$
    – Lee Mosher
    Apr 2, 2021 at 0:42
  • $\begingroup$ But, to warn other potential answerers, you should include in your question any hidden restrictions. $\endgroup$
    – Lee Mosher
    Apr 2, 2021 at 0:43
  • $\begingroup$ Guess it's time to look into singular homology then. Many thanks for the guidance. $\endgroup$ Apr 2, 2021 at 0:44

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