# Using cellular homology in computing singular homology

I have a couple of questions on how cellular homology is used in computing singular homology. Here is what should be a simple example given as an exercise in my class, the questions will follow:

Let $X$ be the 2-dimensional CW-complex obtained from $S^1$ by attaching two 2-cells by maps of degree $a$ and $b$, respectively, where $a$ and $b$ are relatively prime integers. Compute the integral homology of $X$.

Now, I understand that generators of $H_n(X^n,X^{n-1})$ can be identified with the $n$-cells of X, so we have the following chain: $0 \rightarrow H_2(X^2,X^1) \rightarrow H_1(X^1,X^0) \rightarrow H_0(H^0) \rightarrow 0$, with the first group $H_2(X^2,X^1)$ generated by $e^2_1$ and $e^2_2$.

What I don't understand is how the Cellular Boundary Formula: $d_n(e^n_\alpha)=\Sigma_\beta{d_{\alpha\beta}e^{n-1}_\beta}$ is used. Hatcher's book says that $d_{\alpha\beta}$ is the degree of "the composition of the attaching map of the $n$-cell $e^n_\alpha$ with the quotient map collapsing $X^{n-1}-e^{n-1}_\beta$ to a point". In our example above we have the degree of the map from each 2-cell to $S^1$, but we don't know the CW-structure of that $S^1$ (it could be one 1-cell and one 0-cell, or two 1-cells and two 0-cells etc.), so how am I supposed to figure out what to sum over in the formula above to calculate the $d_2$ map? The homology groups must be the same whatever structure of $S^1$ we use, so could I just use any structure? The degree of a composition is the product of the degrees of the components, and we know the degree of one component (those are $a$ and $b$ in the example, right?) but what is the degree of the collapsing map? As you can see I am quite a bit confused about this, no intuition whatsoever, so any help would be greatly appreciated. If I am looking at this completely wrong do correct me, please.

• So just to confirm I understood this: since we have $d_2(e^2_1)=ae^1_2$ and $d_2(e^2_2)=be^1_2$ and $(a,b)=1$ it follows that $Ker(d_2)\cong\mathbb{Z}$, so $H_2(X)\cong\mathbb{Z}$. Also, since we put the one 1-cell one 0-cell structure on $S^1$ we have that $d_1$ is a zero map, $ker(d_1)\cong\mathbb{Z}$ and $H_0(X)\cong\mathbb{Z}$. Finaly, since $H_2(X^2,X^1)\cong\mathbb{Z}\oplus\mathbb{Z}$ and $Ker(d_2)\cong\mathbb{Z}$ then $H_1(X)=0$. Is this right? Feb 28, 2015 at 18:59
• That all looks correct up to the last sentence. The result is correct, but your reasoning isn't clear. Unless I'm horrifically mistaken $H_1(X)\equiv\frac{ker(d_1)}{im(d_2)}$ since $(a,b)=1$, $d_2$ is surjective and therefore $H_1(X)\equiv 0$ Feb 28, 2015 at 19:30
• Why do you say that $H_1(X)\cong\frac{ker(d_1)}{im(d_2)}$ since $(a,b)=1$? Isn't it always equal to that quotient? Or is that comma in the wrong place? :) Feb 28, 2015 at 19:38