[proofplan]
We prove $\chi(X) = \chi_{\mathbb{Z}}(X) = \chi_{\mathbb{F}}(X)$ by working with the cellular chain complex $C_\bullet^{\mathrm{cell}}(X)$. Using the short exact sequences $0 \to Z_n \to C_n \to B_{n-1} \to 0$ and $0 \to B_n \to Z_n \to H_n \to 0$, together with the rank-nullity relation for free abelian groups, we show that the alternating sum of ranks of the chain groups equals the alternating sum of ranks of the homology groups. The field version follows from the universal coefficient theorem.
[/proofplan]
[step:Define cycles, boundaries, and homology and establish the two short exact sequences]
Write $C_n = C_n^{\mathrm{cell}}(X)$ for the cellular chain group in degree $n$. Since $X$ is a finite cell complex, each $C_n$ is a finitely generated free abelian group with $\operatorname{rank}(C_n) = c_n$, the number of $n$-cells. Define
\begin{align*}
Z_n &= \ker(d_n: C_n \to C_{n-1}), \\
B_n &= \operatorname{im}(d_{n+1}: C_{n+1} \to C_n).
\end{align*}
Since $C_n$ is free abelian and $Z_n$ is a subgroup of a finitely generated free abelian group, $Z_n$ is itself free abelian. Similarly, $B_n$ is a subgroup of the free abelian group $Z_n$, hence free abelian. There are two short exact sequences:
\begin{align*}
0 \to Z_n \xrightarrow{\iota} C_n \xrightarrow{d_n} B_{n-1} \to 0, \\
0 \to B_n \xrightarrow{\iota} Z_n \to H_n^{\mathrm{cell}}(X) \to 0,
\end{align*}
where the first sequence uses the fact that $d_n$ is surjective onto its image $B_{n-1}$, and the second is the definition of homology as $H_n^{\mathrm{cell}}(X) = Z_n / B_n$.
[guided]
We work with the cellular chain complex because its chain groups have finite rank equal to the number of cells, making the alternating sum of ranks directly the Euler characteristic $\chi(X) = \sum_n (-1)^n c_n$.
The groups $Z_n$ and $B_n$ are subgroups of finitely generated free abelian groups. By the classification of subgroups of free abelian groups (every subgroup of a finitely generated free abelian group is free abelian), both $Z_n$ and $B_n$ are free abelian of finite rank.
The sequence $0 \to Z_n \to C_n \xrightarrow{d_n} B_{n-1} \to 0$ is exact because $Z_n = \ker(d_n)$ and $d_n$ is surjective onto $B_{n-1} = \operatorname{im}(d_n)$ by definition.
The sequence $0 \to B_n \to Z_n \to H_n^{\mathrm{cell}}(X) \to 0$ is exact because $H_n^{\mathrm{cell}}(X) = Z_n / B_n$.
Both sequences split (the quotients $B_{n-1}$ and $H_n^{\mathrm{cell}}(X)$ are free abelian, hence projective), so the rank is additive across each.
[/guided]
[/step]
[step:Derive the rank identity $c_n = \operatorname{rank}(Z_n) + \operatorname{rank}(B_{n-1})$]
Since $B_{n-1}$ is free abelian, the first short exact sequence splits, giving
\begin{align*}
\operatorname{rank}(C_n) = \operatorname{rank}(Z_n) + \operatorname{rank}(B_{n-1}).
\end{align*}
Similarly, since $H_n^{\mathrm{cell}}(X)$ may have torsion but $B_n$ is free, the rank is still additive: from the second sequence,
\begin{align*}
\operatorname{rank}(Z_n) = \operatorname{rank}(B_n) + \operatorname{rank}(H_n^{\mathrm{cell}}(X)).
\end{align*}
Here $\operatorname{rank}(H_n^{\mathrm{cell}}(X))$ denotes the rank of the free part, i.e., the $n$-th Betti number $\beta_n$.
[/step]
[step:Compute the alternating sum to establish $\chi(X) = \chi_{\mathbb{Z}}(X)$]
Taking the alternating sum:
\begin{align*}
\chi(X) &= \sum_n (-1)^n c_n = \sum_n (-1)^n \bigl(\operatorname{rank}(Z_n) + \operatorname{rank}(B_{n-1})\bigr) \\
&= \sum_n (-1)^n \operatorname{rank}(Z_n) + \sum_n (-1)^n \operatorname{rank}(B_{n-1}).
\end{align*}
Re-indexing the second sum by setting $m = n - 1$:
\begin{align*}
\sum_n (-1)^n \operatorname{rank}(B_{n-1}) = \sum_m (-1)^{m+1} \operatorname{rank}(B_m) = -\sum_m (-1)^m \operatorname{rank}(B_m).
\end{align*}
Substituting $\operatorname{rank}(Z_n) = \operatorname{rank}(B_n) + \beta_n$ from the second short exact sequence:
\begin{align*}
\chi(X) &= \sum_n (-1)^n \bigl(\operatorname{rank}(B_n) + \beta_n\bigr) - \sum_n (-1)^n \operatorname{rank}(B_n) \\
&= \sum_n (-1)^n \beta_n = \chi_{\mathbb{Z}}(X).
\end{align*}
The boundary-rank terms cancel in pairs, leaving the alternating sum of Betti numbers.
[guided]
The computation hinges on a telescoping cancellation. We have
\begin{align*}
\chi(X) = \sum_n (-1)^n c_n = \sum_n (-1)^n \operatorname{rank}(Z_n) + \sum_n (-1)^n \operatorname{rank}(B_{n-1}).
\end{align*}
The second sum, after re-indexing $m = n - 1$, becomes $-\sum_m (-1)^m \operatorname{rank}(B_m)$. Now substitute $\operatorname{rank}(Z_n) = \operatorname{rank}(B_n) + \beta_n$ into the first sum:
\begin{align*}
\chi(X) &= \sum_n (-1)^n \operatorname{rank}(B_n) + \sum_n (-1)^n \beta_n - \sum_n (-1)^n \operatorname{rank}(B_n) \\
&= \sum_n (-1)^n \beta_n = \chi_{\mathbb{Z}}(X).
\end{align*}
Why do the boundary-rank terms cancel? Because the same quantity $\sum_n (-1)^n \operatorname{rank}(B_n)$ appears once with a positive sign (from expanding $\operatorname{rank}(Z_n)$) and once with a negative sign (from re-indexing the $\operatorname{rank}(B_{n-1})$ sum). This is the essential mechanism: the Euler characteristic depends only on the homology, not on the particular chain complex used to compute it.
[/guided]
[/step]
[step:Deduce $\chi_{\mathbb{F}}(X) = \chi_{\mathbb{Z}}(X)$ via the universal coefficient theorem]
By the [Cellular Homology Equals Singular Homology](/theorems/2263) theorem, $H_n^{\mathrm{cell}}(X) \cong H_n(X)$, so $\chi_{\mathbb{Z}}(X) = \sum_n (-1)^n \operatorname{rank}(H_n(X; \mathbb{Z}))$.
The universal coefficient theorem gives
\begin{align*}
H_n(X; \mathbb{F}) \cong \bigl(H_n(X; \mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{F}\bigr) \oplus \operatorname{Tor}_1^{\mathbb{Z}}(H_{n-1}(X; \mathbb{Z}), \mathbb{F}).
\end{align*}
Therefore
\begin{align*}
\dim_{\mathbb{F}} H_n(X; \mathbb{F}) = \beta_n + t_{n-1},
\end{align*}
where $t_m = \dim_{\mathbb{F}} \operatorname{Tor}_1^{\mathbb{Z}}(H_m(X; \mathbb{Z}), \mathbb{F})$. Taking the alternating sum:
\begin{align*}
\chi_{\mathbb{F}}(X) = \sum_n (-1)^n (\beta_n + t_{n-1}) = \sum_n (-1)^n \beta_n + \sum_n (-1)^n t_{n-1}.
\end{align*}
Re-indexing the second sum by $m = n - 1$:
\begin{align*}
\sum_n (-1)^n t_{n-1} = -\sum_m (-1)^m t_m.
\end{align*}
The same argument applied to the chain complex $C_\bullet(X; \mathbb{F})$ shows that $\sum_m (-1)^m t_m$ cancels against its own re-indexed version, giving $\chi_{\mathbb{F}}(X) = \sum_n (-1)^n \beta_n = \chi_{\mathbb{Z}}(X)$.
Alternatively, the same rank-nullity argument from the previous step applies directly to the $\mathbb{F}$-vector space chain complex $C_\bullet^{\mathrm{cell}}(X; \mathbb{F}) = C_\bullet^{\mathrm{cell}}(X) \otimes_{\mathbb{Z}} \mathbb{F}$, yielding $\chi(X) = \sum_n (-1)^n \dim_{\mathbb{F}} C_n^{\mathrm{cell}}(X; \mathbb{F}) = \sum_n (-1)^n \dim_{\mathbb{F}} H_n(X; \mathbb{F}) = \chi_{\mathbb{F}}(X)$.
[/step]
