----- > [!proposition] Proposition. ([[how singular homology interacts with cell complexes]]) > Let $X$ be a [[cell complex]]. Then can say the following about when $i=n$, $i > n$, and $i < n$: >1. $H_{i}(X^{n}, X^{n-1}) \cong \begin{cases} \mathbb{Z}\langle n\text{-cells} \rangle & i=n \\ 0 & \text{else.}\end{cases}$ >2. $H_{i}(X^{n}) \cong 0$ if $i > n$ >3. The inclusion $\iota:X^{n} \xhookrightarrow{}X$ [[singular (co)chain map and homomorphism induced by a continuous map|induces]] an [[isomorphism]] $\iota_{*}:H_{i}(X^{n}) \xrightarrow{\sim} H_{i}(X)$ for $i<n$ > ![[Pasted image 20250506163220.png]] > ^proposition > [!proof]- Proof. ([[how singular homology interacts with cell complexes]]) > For now we only prove $(3)$ assuming $X$ is finite-dimensional. > > > Summary: > 1. $(X,X^{n-1})$ is good, so really are computing $\widetilde{H}_{i}(X^{n} / X^{n-1})$. But $X^{n} / X^{n-1}$ is a wedge of spheres > 2. Long exact sequence for the pair $(X^{n}, X^{n-1})$ > 3. Assume $X=X^{N}$ finite-dimensional. Exact sequence of pair $(X^{n+1}, X^{n})$ to get $H_{i}(X^{n}) \cong H_{i}(X^{n+1})$. Then repeat all the way up till $X^{N}$ > > **1.** $X^{n-1} \subset X^{n}$ is a [[cell complex|subcomplex]], thus $(X, X^{n-1})$ forms a [[good pair]] by [[subcell complexes form good pairs]]. Thus we may invoke [[relative homology for a good pair is reduced homology of the quotient]] to have ${H}_{i}(X^{n},X^{n-1}) \cong \widetilde{H}_{i}(X^{n} / X^{n-1}).$ > But the [[quotient space|space]] $X^{n} / X^{n-1}$ is just a [[wedge sum|wedge of spheres]], one for each $n$-cell. And we know that the [[reduced homology of a good sum of wedges is direct sum of reduced homologies]]; this gives the result. > > **2.** Just consider the [[long exact sequence for relative singular homology|long exact sequence for a pair]] $\dots \to H_{i+1}(X^{n}, X^{n-1}) \xrightarrow{\partial_{}}H_{i}(X^{n-1}) \to H_{i}(X^{n}) \to H_{i}(X^{n}, X^{n-1}) \to \dots$ > where we have just seen in (1) that the first and last groups are zero so long as $i+1 \neq n$ and $i \neq n$, which holds since we have assumed $i>n$. [[exact sequence|Exactness]] implies that the middle map is an [[isomorphism]]. So we get $0=H_{i}(X^{0}) \cong H_{i}(X^{1}) \cong \dots \cong H_{i}(X^{n})$ > and this shows the result. > > **3.** Let $i < n$. For now assume $X$ finite-dimensional, $X=X^{N}$ for some $N$, so $H_{i}(X)=H_{i}(X^{N})$. For $n \leq N$, get an [[exact sequence]] $H_{i+1}(X^{n+1}, X^{n}) \to H_{i}(X^{n}) \to H_{i}(X^{n+1}) \to H_{i}(X^{n+1}, X^{n})$ > by $(1)$, the first and last terms are zero if $i<n$. So $H_{i}(X^{n}) \cong H_{i}(X^{n+1})$. Now inductively obtain $H_{i}(X^{n}) \cong H_{i}(X^{n+1}) \cong H_{i}(X^{n+2}) \cong \dots \cong H_{i}(X^{N})=H_{i}(X).$ > ----- #### - [ ] todo want to actually justify beyond visual argument $X^{n}/X^{n-1}$ is wedge sum of spheres, maybe with some category theory ---- #### 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 > ```