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at the moment I am learning a little bit algebraic topology.

So let $(H_\ast, d_\ast)$ be a ordinary homology theory satisfying the additivy axiom in the sense of the Eilenberg Steenrod axioms.

Let $X$ be a CW complex. Then we can define the cellular chain complex of X associated to the $(H_\ast, d_\ast)$ by $C_n^{cell} (X) = H_n(X^n, X^{n-1})$. The boundary operator is given by the boundary operator of the long exact sequence of the triple $(X^n, X^{n-1}, X^{n-2})$. The homology of this chain complex is called the cellular homology associated to $(H_\ast, d_\ast)$ and is denoted by $H_n^{CW}(X)$.

I already proved the following theorem:

If A is a CW-subcomplex of X, then $C_\ast^{cell}(A) \subseteq C_\ast^{cell}(X)$ is a subcomplex.

Now we have an induced long exakt sequence of chain complexes and define the relative cellular homology $H_n^{CW}(X,A)$ by the homology of the quotient complex.

Now I want to prove, that there is an isomorphism $H_n(X,A) \to H_n^{CW}(X,A)$.

I already understand the absolute case (see e.g. Hatcher, p.140). But I do not know, how to extend this proof to the relative case.

My idea was to look at the long exact sequences an use the five lemma

$$ \begin{matrix} H_n^{CW}(A)& \to & H_n^{CW}(X) & \to & H_n^{CW}(X,A) & \to & H_{n-1}^{CW}(A) & \to & H_{n-1}^{CW}(X)\\ \downarrow & & \downarrow & & & & \downarrow & & \downarrow\\ H_n(A)& \to & H_n(X) & \to & H_n(X,A) & \to & H_{n-1}(A) & \to & H_{n-1}(X) \end{matrix} $$

But I do not know how to define the map in the middle of the diagram above.

Thanks in advance, Felix

Edit due to Connor Malin's Answer:

I was able to prove that $$C^{cell}_{n}(X,A) \cong H_n(X^n/A^n, X^{n-1}/A^{n-1})$$ for all $n \geq 1$ and $$C^{cell}_{0}(X,A) \cong H_0(X^0/A^0, A^0/A^0)$$ and the boundary operator is just the boundery operator from the long exact sequence of the tripel $(X^n/A^n, X^{n-1}/A^{n-1}, X^{n-2}/A^{n-2})$ resp. $(X^1/A^1, X^0/A^0, A^0/A^0)$

Now I want to adapt the absolute case to proof that $H_n(X/A, A/A) \cong H_n^{CW}(X,A)$.

So, I considered an aanalog diagram as in Hatcher (p.139):

Edit due to Connor Malin's Answer:

I was able to prove that $$C^{cell}_{n}(X,A) \cong H_n(X^n/A^n, X^{n-1}/A^{n-1})$$ for all $n \geq 1$ and $$C^{cell}_{0}(X,A) \cong H_0(X^0/A^0, A^0/A^0)$$ and the boundary operator is just the boundery operator from the long exact sequence of the tripel $(X^n/A^n, X^{n-1}/A^{n-1}, X^{n-2}/A^{n-2})$ resp. $(X^1/A^1, X^0/A^0, A^0/A^0)$

Now I want to adapt the absolute case to proof that $H_n(X/A, A/A) \cong H_n^{CW}(X,A)$.

So I considered the following diagram as in Hatcher (p.139):

$$\begin{array}{rclrl} &&&& 0\\ &&& \nearrow\\ && H_n(X^{n+1}/A^{n+1}, A/A) &\cong& H_n(X/A, A/A) \\ &&\nearrow \\ &H_n(X^n/A^n, A/A)\\ \nearrow&& \searrow\\ H_{n+1}(X^{n+1}/A^{n+1}, X^n/A^n) & \rightarrow & H_n(X^n/A^n, X^{n-1}/A^{n-1}) & \rightarrow &H_{n-1}(X^{n-1}/A^{n-1}, X^{n-2}/A^{n-2}) \end{array}$$

But I do not know how to choose ?. I already tried $H_n(X^n/A^n, X^{n-1}/A^{n-1})$ and $H_n(X^n/A^n, A^{n-1}/A^{n-1})$. But both seems not to work, since I was not able to show that $H_n(X^{n+1}/A^{n+1}, X^{n-1}/A^{n-1})$ or $H_n(X^{n+1}/A^{n+1}, A^{n-1}/A^{n-1})$ are isomorphic to $H_n(X/A, A/A)$.

And I also see, why this commutes with cellular maps of pairs of CW-complexes. But why does the connecting morphism commute with $d$, i.e. why does

$$\begin{matrix} H_n(X,A) & \overset{d_n}{\rightarrow} & H_{n-1}(A)\\ \downarrow & & \downarrow\\ H_n^{CW}(X,A) & \overset{d_n^{CW}}{\rightarrow} & H_{n-1}^{CW}(A) \end{matrix}$$

commute?

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    $\begingroup$ The construction of cellular homology (as in Hatcher) does not work for infinite CW-complexes for an arbitrary ordinary homology theory. At some point you have to invoke that the $n$-th homology group of a wedge of $k$-spheres is isomorphic to the sum of $n$-th homology groups of a $k$-sphere.For finite wedges this can be derived from the Eilenberg Steenrod axioms. But for infinite wedges this is an additional axiom (which is satisfied for singular homology). $\endgroup$
    – Paul Frost
    Jan 8, 2020 at 13:03
  • $\begingroup$ Hi Paul Frost, thanks for your remark. I just forgot to mention that i assumed also the additivy axiom. I just corrected my post above. $\endgroup$
    – felix
    Jan 11, 2020 at 14:03

1 Answer 1

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From the identification of $X^n / X^{n-1}$ with a wedge of spheres, you can see that your relative chain complex $C^{cell}(X,A)$ is exactly $C^{cell}(X/A)$ with the exception of ${C_0}^{cell}$ which is generated by all the cells except the point $A$ was quotiented to. This means that ${H_*}^{CW}(X,A) \cong {\bar{H}_*}(X/A)$

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  • $\begingroup$ Hi Connor Malin, thank you very much for your answer. This helped a lot. But I have some further questions. I will add them in my question above. $\endgroup$
    – felix
    Jan 11, 2020 at 14:06

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