---- > [!theorem] Theorem. ([[cellular and singular homology agree]]) > For $X$ a [[cell complex]], we have a [[natural transformation|natural]] [[isomorphism]] $H_{*}^{\text{cell}}(X)\cong H_{*}(X)$ between its [[cellular homology]] and [[singular homology]]; [[singular cohomology|cohomology]] too. ^theorem > [!proposition] Corollary. > 1. If $X$ is a finite [[cell complex]], then $H_{n}(X)$ is a [[submodule generated by a subset|finitely generated module]] for all $n$ (as a [[quotient module|quotient]] thereof), generated by at most $|I_{n}|$ elements. In particular, if there are no $n$-cells, $H_{n}(X)=0$. > 2. Building on this: if $X$ has only even-index cells (e.g. $\mathbb{S}^{2}$ or $\mathbb{C}P^{k}$) then $H_{n}(X)=C_{n}^{\text{cell}}(X)$ — no differential computation required. ^proposition > [!proof]- Proof. ([[cellular and singular homology agree]]) > > Summary: draw the diagram with tan differentials triangling up and down, then fill in with LES of a pair to get SES on the diagonals. Then chase. > > Fix $n$. Much of the work is baked into [[how singular homology interacts with cell complexes]]. Start with the composition $H_{n+1}(X^{n+1}, X^{n}) \xrightarrow{d_{n+1}^{\text{cell}}} H_{n}(X^{n}, X^{n-1}) \xrightarrow{d_{n}^{\text{cell}}}H_{n-1}(X^{n-1}, X^{n-2}).$ > Recalling how $d_{n}^{\text{cell}}$ is defined and employing [[long exact sequence for relative singular homology|long exact sequence for a pair]] liberally, get a diagram > > ![[Pasted image 20250518202207.png|500]] > where the diagonals are [[exact sequence|exact]]. Note the tan compositions are the cellular boundary maps $d$. By [[how singular homology interacts with cell complexes]], know $H_{n}(X) \cong H_{n}(X^{n+1})$. And exactness then gives $H_{n}(X^{n+1}) \cong \operatorname{coker }\partial_{}= H_{n}(X^{n}) / (\operatorname{im}(\partial_{}: H_{n+1}(X^{n+1}, X^{n}) \to H_{n}(X^{n})))$ > Now, $q_{n}$ as pictured is [[injection|injective]] (by short-exactness), hence an isomorphism onto its image, so we may apply it to top and bottom here to get $\cong \frac{q_n\big( H_{n}(X^{n}) \big)}{d_{n}^{\text{cell}}: H_{n+1}(X^{n+1}, X^{n}) \to H_{n}(X^{n}, X^{n-1})}.$ > By exactness, the image of $q_{n}$ is the kernel of $\partial_{}$. So we have $\begin{align} > &= \frac{\operatorname{ker}(\partial: H_{n}(X^{n}, X^{n-1}) \to H_{n-1}(X^{n-1}))}{\operatorname{im }d_{n}^{\text{cell}}} \\ > &=\frac{\operatorname{ker }d_{n}^{\text{cell}}}{\operatorname{im }d_{n+1}^{\text{cell}}} \\ > &= H_{n}^{\text{cell}}(X). > \end{align}$ > > > ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```