$\chi(X)=\chi(A)+\chi(X/A)-1$

Question

Let $X$ be a finite CW complex, and $A$ a subcomplex. What I want to show is $\chi(X)=\chi(A)+\chi(X/A)-1$. Of course I can show this by just counting, that for $n>0$, (the number of $n$-cells of $X$) equals the sum of (the number of $n$-cells of $A$)$+$(the number of $n$-cells of $X/A$), and for $n=0$, (the number of $0$-cells of $A$)+(then number of $0$-cells of $X/A$)$=$(the number of $0$-cells of $X$)$+1$. But I want to prove this more rigorously.

I know that $\chi(X)=\sum_{n \geq 0} (-1)^n$rank $H_n(X) $. Can I use this to prove $\chi(X)=\chi(A)+\chi(X/A)-1$ by means of some long exact sequences, or etc?

Answer 1

Note that since $X$ is finite, its cellular homology chain complex has finite length,

$$ \newcommand{\im}{\operatorname{im}} \newcommand{\rk}{\operatorname{rk}} 0 \to C_n(X) \to \cdots \to C_0(X) \to 0. $$

Now, it is a general fact for finite length chain complexes that

$$ \sum_{i}(-1)^i\rk C_i = \sum_{i}(-1)^i\rk H_i(C_\bullet). $$

To see this, from the exact sequences

$$ 0 \to \im d_{q+1} \to \ker d_q \to \ker d_q/\im d_{q+1} = H_q(C_\bullet) \to 0 $$

and

$$ 0 \to \ker d_q \to C_q \to \im d_q \to 0 $$

we have

$$ \rk C_q = \rk \ker d_q + \rk \im d_q = \rk H_q(C_\bullet) + \rk \im d_{q+1} + \rk \im d_q. $$

It suffices now to use this equality and expand the alternating sum.

Going back to the question, since

$$ 0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X,A) \to 0 $$

is exact we have short exact sequences $0 \to C_q(A) \to C_q(X) \to C_q(X,A) \to 0$, which give

$$ \rk C_q(X) = \rk C_q(X) + \rk C_q(X,A). $$

By passing to the alternating sum of ranks on the cellular complexes, we get

$$ \begin{align} \chi(X) &= \sum_i(-1)^i\rk H_i(X) = \sum_i(-1)^i\rk C_i(X)\\ & = \sum_i(-1)^i\rk C_i(A) + \sum_i(-1)^i\rk C_i(X,A)\\ &= \sum_i(-1)^i\rk H_i(A) + \sum_i(-1)^i\rk H_i(X,A)\\ & = \chi(A) + \sum_i(-1)^i\rk H_i(X,A), \end{align} $$

and since $A$ is a subcomplex, by excision we know that

$$ H_i(X,A) \simeq \widetilde{H}_i(X/A). $$

Finally, from the fact that $H_i(X/A) = \widetilde{H}_i(X/A)$ and $H_0(X/A) \oplus \mathbb{Z} = \widetilde{H}_0(X/A)$ we obtain

$$ \rk H_i(X,A) = \rk \widetilde{H}_i(X/A) = \rk H_i(X/A) - \delta_{i0} $$

where $\delta$ is the Kroenecker delta, and plugging this into the alternating sum yields

$$ \chi(A) = \chi(X) + \chi(X/A) - 1 $$

as desired.