---- > [!theorem] Theorem. ([[the excision theorem]]) > Let $(X,A)$ be a [[topological pair]] and $Z \subset A$ be such that its [[closure]] lies in the [[topological interior|interior]] of $A$: $\overline{Z} \subset \text{int }A$. Then the [[inclusion map|inclusion]] [[topological pair|map of pairs]] $(X - Z, A - Z) \xhookrightarrow{} (X,A)$ [[relative singular homology|induces]] [[isomorphism|isomorphisms]] $H_{n}(X - Z, A - Z) \xrightarrow{\cong} H_{n}(X,A)$ on [[relative singular homology]]. ^theorem > [!intuition] > The main point of [[relative singular homology]] is that we want to think of $H_{n}(X,A)$ as the '[[(co)homology of a complex|homology]] of $X$ when we ignore $A. Thus, one might expect that the [[relative homology of an embedding of chain complexes|relative homology]] does not depend on things 'inside $A. However, it is not true in general that, say, $H_{n}(X,A)=H_{n}(X-A)$. Instead, what are allowed to do is remove — to *excise* — [[subspace topology|subspaces]] of $A$ that are 'not too big'. This is the content of the excision theorem. ^intuition > [!proposition] Corollary. (Excise $X-A$) > $H_{n}(X, X-Z) \xleftarrow{\cong} H_{n}(A, A-Z)$ . > > Indeed, $\begin{align} H_{n}(X, X-Z) & \xleftarrow{\cong} H_{n}(X - (X - A), X-Z - (X-A)) \text{ (excision)} \\ &= \text{H}_{n}(A, A-Z). \end{align}$ ^proposition - [ ] statement for cohomology (note contravariance) > [!proof]- Proof. ([[the excision theorem]]) > Let $X \supset A \supset Z$ be such that > Summary: > 1. Pick $\mathcal{U}=\{ A,B \}$, with $B=X-Z$. Write $C_{\bullet}(A+B):=C_{\bullet}^{ \mathcal{U}}(X)$ > 2. Get an inclusion of chain complex SES, one corresponding to $\mathcal{U}$-chains and the one without. In the middle will be $C_{\bullet}(A+B) \hookrightarrow C_{\bullet}(X)$ > 3. Naturality of LES on homology gives a diagram with $H_{i}^{\mathcal{U}}(X,A) \to H_{i}(X,A)$ in the middle. Argue with [[small simplices theorem]], then [[five lemma]], that this map is an [[isomorphism]] > 4. Now, by definition $H_{i}^{\mathcal{U}}(X,A)=H_{i}\big( \frac{C_{\bullet}^{\mathcal{U}}(X)}{C_{\bullet}(A)} \big)$. Argue that $\frac{C_{\bullet}(B)}{C_{\bullet}(A \cap B)} \to \frac{C_{\bullet}(A+B)}{C_{\bullet}(A)}$ is an [[isomorphism]]. The point is that $A \cap B$ is closely related to $A-Z$. Then finish. > > The assumption $\overline{Z} \subset \text{int }A$ will allow us to employ the [[small simplices theorem]], from which the result will follow. Another result that we don't explicitly use, but take a similar strategy to at one part, is [[detecting isomorphisms with relative homology]]. > > Let $X \supset A \supset Z$ be such that $\overline{Z} \subset \text{int }A$. Put $B=X-Z$. Then, as in the proof of [[Mayer-Vietoris theorem|Mayer-Vietoris]], take $\mathcal{U}=\{ A,B \}$. By assumption, the interiors cover $X$.[^1] > > As in the [[Mayer-Vietoris theorem|Mayer-Vietoris]] proof, write $C_{\bullet}(A+B)=C_{\bullet}^{\mathcal{U}}(X)$, where $C_{\bullet}^{\mathcal{U}}(X)$ is as in the [[small simplices theorem]]. We consider the [[short exact sequence|short exact sequences]] of [[chain complex of modules|chain complexes]] > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRGJAF9T1Nd9CKAIzkqtRizYBhAPoAdOQCMmDBjBwAKAIIBKLjxAZseAkQBMo6vWatEIWQuWr12gNQAhPd17GBRAMyW4jbS8koqappaHjoABAD0sQ7hzlFeBkb8pigALEHWknYc3oZ8JoIkpEJiBbbs+j5ZFSLVVhJ1yU6R2umN5eZVNe2hjhEuABq9pb7ZyIGtwYX2YV0TpLoN001EeQu1bMViMFAA5vBEoABmAE4QALZIZCA4EEhCJTf3b9QvSGYftweiAsz1eiH8AK+4J+YJykKBAFYYUgAGzw1HIxAAdnR2MxAA5cfjMQBOXEiUEYkAMLBgOpQCBMRRqEDUAAWMDoUDYkDprOedCwDB5BFYuJBv2xuMClMQhIonCAA > \begin{tikzcd} > 0 \arrow[r] & C_\bullet(A) \arrow[r] \arrow[d, no head, Rightarrow] & C_\bullet(A+B) \arrow[r] \arrow[d] & C_\bullet(A+B) / C_\bullet(A) \arrow[r] \arrow[d] & 0 \\ > 0 \arrow[r] & C_\bullet(A) \arrow[r] & C_\bullet(X) \arrow[r] & {C_\bullet(X,A)} \arrow[r] & 0 > \end{tikzcd} > \end{document} > ``` > > where the vertical maps are [[inclusion map|inclusions]]. The middle map [[singular (co)chain map and homomorphism induced by a continuous map|induces]] an [[isomorphism]] on [[(co)homology of a complex|homology]], by the [[small simplices theorem]]. Naturality of the [[long exact sequence on homology induced by short exact sequence of chain complexes|LES on homology]] (or something like that) gives us a diagram (hopefully the notation is obvious) > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB12BjKCHBAL6l0mXPkIoAjOSq1GLNgAkA+lgAUAQQCUIISOx4CRAEwzq9Zq0QgV6gBo69IDAfFEAzGbmWlqtXdJtXWFnUUMJZAAWLwsFaxVgLABaSQFNRxCXMSMUAFYY+SsbZUSUtIdg-WyIgDYCn2tOHj5BTLC3KVJJWVii2zUAIQyq8JMunsLfdQBNYdDXHORPbvNJ+L9p0iHK+eqiaJXvOOLS1MG5rNG88dWGk+Sz2Z3LjuQ6w962Jt5+Z-bFsgfNYcbg-QSyGBQADm8CIoAAZgAnCAAWyQZBAOAgSFSISRqJx1CxSGMTnxaMQpkx2MQ7jJyIpnmpSGiRyKnDQdEReEYO3JLKJNNy9IJiHyzMQNRFFMkTOJiAA7NKkArBUgABzKxDqtWIACcWr1uskGM+jXYnO5WF51AYdAARjAGAAFf4SEAMGDwnB8hk4jHyyS4hF+xDSCVKvGhqnyzVR0VymkG+MU1mB4ha8WB4Mgflh8OB0kUARAA > \begin{tikzcd} > \cdots \arrow[r] & H_i(A) \arrow[r] \arrow[d] & H_i(A+B) \arrow[r] \arrow[d] & {H_i^\mathcal{U}(X,A)} \arrow[r, "\partial"] \arrow[d] & H_{i-1}(A) \arrow[r] \arrow[d] & H_{i-1}(A+B) \arrow[r] \arrow[d] & \cdots \\ > \cdots \arrow[r] & H_i(A) \arrow[r] & H_i(X) \arrow[r] & {H_i(X,A)} \arrow[r, "\partial"] & H_{i-1}(A) \arrow[r] & H_{i-1}(X) \arrow[r] & \cdots > \end{tikzcd} > \end{document} > ``` > Where all maps but the middle are isomorphisms. The [[five lemma]] now says the middle map is an isomorphism: $H_{i}\left( \frac{C_{\bullet}(A+B)}{C_{\bullet}^{\mathcal{U}}(A)} \right)\cong H_{i}(X,A).$ > On the other hand, the map $\frac{C_{\bullet}(B)}{C_{\bullet}(A \cap B)} \to \frac{C_{\bullet}(A+B)}{C_{\bullet}(A)}$is an isomorphism of [[chain complex of modules|chain complexes]]. Passing to homology gives us an isomorphism $H_{*}(B, A \cap B) \cong H_{*}(X,A)$. Substituting $B=X-Z$, we obtain $H_{*}(X-Z, A \cap (X-Z)) \cong H_{*}(X, A).$ > Since $Z \subset \overline{Z} \subset \text{int }A$, $A\cap (X-Z)=A-Z$, giving us the result. ---- #### [^1]: Recall from [[closure and interior]] that $\text{int}(X-Z)=X-\overline{Z}$. So $\text{int }A \cup \text{int}(X-Z)=\text{int }A \cup (X-\overline{Z})$. Since $\overline{Z} \subset \text{int }A$, $X-\overline{Z} \supset X-\text{int }A$. Hence $\text{int }A \cup \text{int}(X-Z) \supset \text{int }A \cup \text{int}(X-A)=X$. ----- #### 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 > ```