# Isomorphism of the homology group to the associated cellular homology group

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.

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?

• 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). Jan 8, 2020 at 13:03
• 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. Jan 11, 2020 at 14:03

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 $${H_*}^{CW}(X,A) \cong {\bar{H}_*}(X/A)$$