---- Let - $X$ be a [[topological space]], - $A \xrightarrow{\iota} X$ a [[subspace topology|subspace]], - $C_{\bullet}(X),C_{\bullet}(A), C_{\bullet}(X,A)$ the corresponding [[singular homology|singular]] [[chain complex of modules|chain complexes]], and - $H_{\bullet}(X),H_{\bullet}(A),H_{\bullet}(X,A)$ the corresponding [[singular homology|singular]] [[(co)homology of a complex|homology]]/[[relative singular homology|relative homology]]. > [!theorem] Theorem. ([[long exact sequence for relative singular homology]]) > **1.** There are [[linear map|homomorphisms]] $\partial:H_{n}(X,A) \to H_{n-1}(A)$, given by mapping[^1] $\big[ [c] \big] \mapsto [d_{n}c],$ > which fit into a [[exact sequence|long exact sequence]] > > ![[CleanShot 2025-04-01 at [email protected]]] > > where $\iota_{*}$ is [[singular (co)chain map and homomorphism induced by a continuous map|induced]] by the [[inclusion map|inclusion map]] $\iota:A \hookrightarrow X$ and $q_{*}$ is [[homomorphism on homology induced by a chain map|induced by]] the [[quotient group|quotient homomorphism]] $C_{\bullet}(X) \to C_{\bullet}(X,A)$. This sequence is called the **long exact sequence for the [[topological pair|pair]]** $(X,A)$. > > **2.** This sequence is natural (i.e., compatible) with respect to any [[topological pair|map of pairs]] $(X,A) \to (Y,B)$. > [!proof]+ Proof. ([[long exact sequence for relative singular homology]]) > This result is actually just homological algebra, and in particular follows from [[the snake lemma]]. Specifically, since $\iota_{n}$ is [[injection|injective]], $q_{n}:C_{n}(X) \to C_{n}(X) / C_{n}(A)=C_{n}(X,A)$ is [[surjection|surjective]], and $\operatorname{im }\iota_{n} = C_{n}(A)=\operatorname{ker }q_{n}$, in each grading we have a [[short exact sequence]] $0 \to C_{n}(A) \to C_{n}(X) \to C_{n}(X,A) \to 0$. Now the [[long exact sequence on homology induced by short exact sequence of chain complexes]], a corollary of [[the snake lemma]], applies to give the result. ---- #### [^1]: This makes sense because if $c \in C_{n}(X)$, then $[c] \in \frac{C_{n}(X)}{C_{n}(A)}$. Since $[c]$ is a (relative) cycle, we know $\overline{d_{n}}(c)=[d_{n}c]=0$. Hence $d_{n}c \in C_{n-1}(A)$. Since $d^{2}=0$, $d_{n}c \in \operatorname{ker }d_{n}$ is a cycle, so indeed it makes sense to talk about $[d_{n}c]$ as an element of $H_{n-1}(A)$. ----- #### 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 > ```