[proofplan] We set $B = X \setminus Z$ and verify that $\mathcal{U} = \{A, B\}$ is a valid open cover of $X$ (i.e., $\mathring{A} \cup \mathring{B} = X$). We then construct a commutative diagram relating two short exact sequences of chain complexes: the top row $0 \to C_\bullet(A) \to C_\bullet^{\mathcal{U}}(X) \to C_\bullet^{\mathcal{U}}(X)/C_\bullet(A) \to 0$ and the bottom row $0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X, A) \to 0$. The [Small Simplices Theorem](/theorems/2255) shows the middle vertical inclusion induces isomorphisms on homology, and the [Five Lemma](/theorems/1938) then forces the right vertical map to induce isomorphisms as well. Finally, we identify the quotient $C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)$ with the relative chain complex $C_\bullet(B, A \cap B) = C_\bullet(X \setminus Z, A \setminus Z)$, completing the proof. [/proofplan] [step:Set $B = X \setminus Z$ and verify that $\mathring{A}$ and $\mathring{B}$ cover $X$] Define $B = X \setminus Z$. Since $Z$ is a subset of $X$, the set $B$ is the complement of $Z$ in $X$. The interior of $B$ in $X$ is $\mathring{B} = X \setminus \overline{Z}$: a point $x \in B$ has a neighbourhood contained in $B$ if and only if $x \notin \overline{Z}$. We verify that $\mathring{A} \cup \mathring{B} = X$. Let $x \in X$. Either $x \notin \overline{Z}$, in which case $x \in X \setminus \overline{Z} = \mathring{B}$, or $x \in \overline{Z}$. In the latter case, the hypothesis $\overline{Z} \subset \mathring{A}$ gives $x \in \mathring{A}$. Therefore $\mathring{A} \cup \mathring{B} = X$, and $\mathcal{U} = \{A, B\}$ is a valid open cover of $X$ in the sense required by the [Small Simplices Theorem](/theorems/2255). [guided] The hypothesis $\overline{Z} \subset \mathring{A}$ is doing two things simultaneously. First, it ensures $Z \subset A$ (which we already know). Second, and more importantly, it ensures that the interiors of $A$ and $B$ cover $X$. Without this hypothesis, the cover $\{A, B\}$ might fail to have the property $\mathring{A} \cup \mathring{B} = X$. For example, if $Z = A$ and $A$ is closed with empty interior, then $\mathring{A} = \varnothing$ and $\mathring{B} = X \setminus A$, which misses boundary points of $A$. Why do we need the interiors to cover $X$? The [Small Simplices Theorem](/theorems/2255) requires an open cover $\{U_\lambda\}$ such that $X = \bigcup_\lambda U_\lambda$. We do not have $A$ and $B$ open in general, but the theorem applies with $\mathcal{U} = \{A, B\}$ as long as $\mathring{A} \cup \mathring{B} = X$, because the Lebesgue number argument in the proof of the Small Simplices Theorem uses the interiors of the cover elements. [/guided] [/step] [step:Construct the commutative diagram of short exact sequences] Write $C_\bullet^{\mathcal{U}}(X) = C_\bullet(A + B)$ for the subcomplex of $C_\bullet(X)$ consisting of chains each of whose singular simplices has image contained in $A$ or in $B$. The inclusion $C_\bullet(A) \hookrightarrow C_\bullet^{\mathcal{U}}(X)$ is well-defined since $A$ is one of the cover elements, and the quotient $C_\bullet^{\mathcal{U}}(X) / C_\bullet(A)$ is a chain complex with the induced differential. Consider the two short exact sequences of chain complexes: \begin{align*} 0 \to C_\bullet(A) \xrightarrow{\iota} C_\bullet^{\mathcal{U}}(X) \xrightarrow{q} C_\bullet^{\mathcal{U}}(X)/C_\bullet(A) \to 0 \end{align*} \begin{align*} 0 \to C_\bullet(A) \xrightarrow{i} C_\bullet(X) \xrightarrow{p} C_\bullet(X)/C_\bullet(A) \to 0 \end{align*} where $C_\bullet(X)/C_\bullet(A) = C_\bullet(X, A)$ is the relative chain complex. These fit into a commutative diagram with vertical maps given by the natural inclusions: \begin{align*} \begin{array}{ccccccccc} 0 & \to & C_\bullet(A) & \xrightarrow{\iota} & C_\bullet^{\mathcal{U}}(X) & \xrightarrow{q} & C_\bullet^{\mathcal{U}}(X)/C_\bullet(A) & \to & 0 \\ & & \| & & \downarrow \alpha & & \downarrow \beta & & \\ 0 & \to & C_\bullet(A) & \xrightarrow{i} & C_\bullet(X) & \xrightarrow{p} & C_\bullet(X, A) & \to & 0 \end{array} \end{align*} The left vertical map is the identity on $C_\bullet(A)$. The middle vertical map $\alpha: C_\bullet^{\mathcal{U}}(X) \hookrightarrow C_\bullet(X)$ is the inclusion of $\mathcal{U}$-small chains into all chains. The right vertical map $\beta: C_\bullet^{\mathcal{U}}(X)/C_\bullet(A) \to C_\bullet(X, A)$ is the map induced on quotients by $\alpha$. We verify exactness of each row. The bottom row is exact by definition of relative chains: $i$ is injective (singular simplices into $A$ are singular simplices into $X$), $p$ is the quotient map hence surjective, and $\ker(p) = \operatorname{im}(i)$. The top row is exact by the same reasoning with $C_\bullet^{\mathcal{U}}(X)$ in place of $C_\bullet(X)$: $\iota$ is injective, $q$ is surjective, and $\ker(q) = \operatorname{im}(\iota)$. We verify commutativity. The left square commutes: $\alpha \circ \iota = i \circ \operatorname{id}$ because both composites send a chain in $C_\bullet(A)$ to the same chain viewed in $C_\bullet(X)$. The right square commutes: for $[c] \in C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)$ represented by $c \in C_\bullet^{\mathcal{U}}(X)$, we have $\beta([c]) = [c]$ in $C_\bullet(X, A)$ and $p(\alpha(c)) = [c]$ in $C_\bullet(X, A)$, so $\beta \circ q = p \circ \alpha$. [guided] Why two rows instead of one? The long exact sequence of a pair $(X, A)$ comes from the bottom row alone (this is the content of the [Long Exact Sequence of a Pair](/theorems/2249)). But the bottom row knows nothing about the cover $\mathcal{U}$. The top row is the version of the same construction using only $\mathcal{U}$-small chains. By comparing the two rows, we will be able to transfer information from the $\mathcal{U}$-small world (where we can identify the quotient explicitly) to the full chain complex world. The key observation is that all three vertical maps are chain maps (they commute with the boundary operator $d$), so the [Snake Lemma](/theorems/1930) applied to each row produces long exact sequences in homology, and the vertical maps induce a morphism between these two long exact sequences. This morphism of long exact sequences is the input to the Five Lemma. [/guided] [/step] [step:Apply the Small Simplices Theorem to show $\alpha_*$ is an isomorphism] The middle vertical map $\alpha: C_\bullet^{\mathcal{U}}(X) \hookrightarrow C_\bullet(X)$ is the inclusion of $\mathcal{U}$-small chains. By the [Small Simplices Theorem](/theorems/2255), the induced map on homology \begin{align*} \alpha_*: H_n^{\mathcal{U}}(X) \xrightarrow{\;\sim\;} H_n(X) \end{align*} is an isomorphism for all $n$. The hypotheses of the Small Simplices Theorem are satisfied: $\mathcal{U} = \{A, B\}$ is a cover of $X$ with $\mathring{A} \cup \mathring{B} = X$ (verified in the first step). [guided] The [Small Simplices Theorem](/theorems/2255) states that for any cover $\mathcal{U}$ of $X$ such that the interiors of the cover elements cover $X$, the inclusion $C_\bullet^{\mathcal{U}}(X) \hookrightarrow C_\bullet(X)$ induces an isomorphism on homology. The proof uses iterated barycentric subdivision: given any singular chain, one can subdivide it sufficiently many times so that each simplex in the subdivision has image contained in some element of $\mathcal{U}$. The chain homotopy between the identity and the subdivision operator (constructed from the [Properties of Barycentric Subdivision](/theorems/???)) ensures that the subdivision represents the same homology class. This is the deepest ingredient in the proof of excision. It is where the topology enters: the compactness of $\Delta^n$ (via the Lebesgue number lemma) guarantees that finitely many subdivisions suffice. [/guided] [/step] [step:Apply the Five Lemma to deduce that $\beta_*$ is an isomorphism] The [Snake Lemma](/theorems/1930) applied to the two short exact sequences produces a morphism of long exact sequences in homology: \begin{align*} \begin{array}{ccccccc} \cdots \to & H_n(A) & \xrightarrow{\iota_*} & H_n^{\mathcal{U}}(X) & \xrightarrow{q_*} & H_n(C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)) & \xrightarrow{\partial} & H_{n-1}(A) & \to \cdots \\ & \| & & \downarrow \alpha_* & & \downarrow \beta_* & & \| & \\ \cdots \to & H_n(A) & \xrightarrow{i_*} & H_n(X) & \xrightarrow{p_*} & H_n(X, A) & \xrightarrow{\partial} & H_{n-1}(A) & \to \cdots \end{array} \end{align*} The vertical maps $H_n(A) \to H_n(A)$ are the identity (isomorphisms), and the vertical maps $\alpha_*: H_n^{\mathcal{U}}(X) \to H_n(X)$ are isomorphisms by the Small Simplices Theorem. By the [Five Lemma](/theorems/1938), applied to five consecutive terms centred on $\beta_*$ in each degree $n$, the map \begin{align*} \beta_*: H_n(C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)) \xrightarrow{\;\sim\;} H_n(X, A) \end{align*} is an isomorphism for all $n$. The [Five Lemma](/theorems/1938) requires that four of the five vertical maps are isomorphisms; in our diagram, the maps on $H_n(A)$, $H_n^{\mathcal{U}}(X)$, $H_{n-1}(A)$, and $H_{n-1}^{\mathcal{U}}(X)$ are all isomorphisms (the first and third are identities; the second and fourth are instances of $\alpha_*$). Therefore $\beta_*$ is an isomorphism. [guided] The Five Lemma is applied to the following five-term segment of the ladder, for each $n$: \begin{align*} \begin{array}{ccccccccc} H_n(A) & \xrightarrow{\iota_*} & H_n^{\mathcal{U}}(X) & \xrightarrow{q_*} & H_n(C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)) & \xrightarrow{\partial} & H_{n-1}(A) & \xrightarrow{\iota_*} & H_{n-1}^{\mathcal{U}}(X) \\ \downarrow \operatorname{id} & & \downarrow \alpha_* & & \downarrow \beta_* & & \downarrow \operatorname{id} & & \downarrow \alpha_* \\ H_n(A) & \xrightarrow{i_*} & H_n(X) & \xrightarrow{p_*} & H_n(X, A) & \xrightarrow{\partial} & H_{n-1}(A) & \xrightarrow{i_*} & H_{n-1}(X) \end{array} \end{align*} The [Five Lemma](/theorems/1938) states: given a morphism of exact sequences $A_1 \to A_2 \to A_3 \to A_4 \to A_5$ and $B_1 \to B_2 \to B_3 \to B_4 \to B_5$ with vertical maps $f_1, f_2, f_3, f_4, f_5$, if $f_1, f_2, f_4, f_5$ are isomorphisms, then $f_3$ is an isomorphism. In our case, $f_1 = \operatorname{id}$, $f_2 = \alpha_*$, $f_3 = \beta_*$, $f_4 = \operatorname{id}$, $f_5 = \alpha_*$, and the four outer maps are all isomorphisms (identities and instances of the Small Simplices isomorphism). The conclusion is that $\beta_*$ is an isomorphism. Note that the Five Lemma needs both rows to be exact. The bottom row is exact by the [Long Exact Sequence of a Pair](/theorems/2249). The top row is exact by the same argument (the Snake Lemma applied to the short exact sequence of chain complexes $0 \to C_\bullet(A) \to C_\bullet^{\mathcal{U}}(X) \to C_\bullet^{\mathcal{U}}(X)/C_\bullet(A) \to 0$). [/guided] [/step] [step:Identify $C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)$ with $C_\bullet(B, A \cap B) = C_\bullet(X \setminus Z, A \setminus Z)$] By definition, $C_n^{\mathcal{U}}(X) = C_n(A) + C_n(B)$ inside $C_n(X)$: a chain is $\mathcal{U}$-small if and only if it is a sum of a chain supported in $A$ and a chain supported in $B$. (Here "supported in $A$" means every singular simplex in the chain has image contained in $A$.) We construct an isomorphism of chain complexes \begin{align*} \varphi: C_\bullet(B) / C_\bullet(A \cap B) \xrightarrow{\;\sim\;} C_\bullet^{\mathcal{U}}(X) / C_\bullet(A). \end{align*} Define $\varphi$ as follows. A class $[\sigma] \in C_n(B)/C_n(A \cap B)$ is represented by a chain $\sigma \in C_n(B) \subset C_n^{\mathcal{U}}(X)$. Map it to the class $[\sigma] \in C_n^{\mathcal{U}}(X)/C_n(A)$. **Well-definedness:** if $\sigma \in C_n(A \cap B)$, then $\sigma \in C_n(A)$ (since $A \cap B \subset A$), so $[\sigma] = 0$ in $C_n^{\mathcal{U}}(X)/C_n(A)$. Hence $\varphi$ is well-defined. **Surjectivity:** let $[c] \in C_n^{\mathcal{U}}(X)/C_n(A)$. Write $c = a + b$ with $a \in C_n(A)$ and $b \in C_n(B)$. Then $[c] = [a + b] = [b]$ in $C_n^{\mathcal{U}}(X)/C_n(A)$, which is in the image of $\varphi$. **Injectivity:** suppose $\varphi([\sigma]) = 0$, i.e., $\sigma \in C_n(A)$. Since $\sigma \in C_n(B)$ by hypothesis, we have $\sigma \in C_n(A) \cap C_n(B) = C_n(A \cap B)$, so $[\sigma] = 0$ in $C_n(B)/C_n(A \cap B)$. The equality $C_n(A) \cap C_n(B) = C_n(A \cap B)$ holds because the singular chain groups are free abelian on their generators: a singular simplex $\sigma: \Delta^n \to X$ lies in both $C_n(A)$ and $C_n(B)$ if and only if $\sigma(\Delta^n) \subset A$ and $\sigma(\Delta^n) \subset B$, i.e., $\sigma(\Delta^n) \subset A \cap B$, i.e., $\sigma \in C_n(A \cap B)$. The map $\varphi$ is a chain map (it commutes with the boundary operator) because the boundary of a chain in $C_n(B)$ lies in $C_{n-1}(B)$, and the quotient maps are compatible with the differentials. Now substituting $B = X \setminus Z$ and $A \cap B = A \cap (X \setminus Z) = A \setminus Z$: \begin{align*} C_\bullet(B)/C_\bullet(A \cap B) = C_\bullet(X \setminus Z)/C_\bullet(A \setminus Z) = C_\bullet(X \setminus Z, A \setminus Z). \end{align*} [guided] The identification $C_n(A) \cap C_n(B) = C_n(A \cap B)$ is the step that uses the freeness of the singular chain groups in an essential way. The singular chain group $C_n(X)$ is the free abelian group on the set of all continuous maps $\Delta^n \to X$. Each generator (a singular simplex $\sigma$) has a definite image $\sigma(\Delta^n)$, and $\sigma$ belongs to $C_n(A)$ if and only if $\sigma(\Delta^n) \subset A$. Since different generators are linearly independent, a chain $\sum_j n_j \sigma_j$ lies in $C_n(A) \cap C_n(B)$ if and only if each $\sigma_j$ with $n_j \neq 0$ satisfies $\sigma_j(\Delta^n) \subset A$ and $\sigma_j(\Delta^n) \subset B$, i.e., $\sigma_j(\Delta^n) \subset A \cap B$. Therefore the chain lies in $C_n(A \cap B)$. Why is this identification important? Without it, we would only know that $C_n(A \cap B) \subset C_n(A) \cap C_n(B)$, and the map $\varphi$ might fail to be injective. The reverse inclusion requires that the only way a formal sum of simplices can lie in both $C_n(A)$ and $C_n(B)$ is if each simplex individually lies in $A \cap B$. This is exactly what freeness guarantees: there are no non-trivial relations among the generators that could allow cancellation to produce an element of $C_n(A) \cap C_n(B)$ from simplices not all in $A \cap B$. [/guided] [/step] [step:Combine the isomorphisms to obtain the excision isomorphism] Composing the isomorphisms from the preceding steps, the inclusion of pairs $(X \setminus Z, A \setminus Z) \hookrightarrow (X, A)$ induces, for each $n$, the chain of isomorphisms \begin{align*} H_n(X \setminus Z, A \setminus Z) \xrightarrow[\varphi_*]{\;\sim\;} H_n(C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)) \xrightarrow[\beta_*]{\;\sim\;} H_n(X, A). \end{align*} The first isomorphism is $\varphi_*$, induced by the chain isomorphism $\varphi: C_\bullet(X \setminus Z, A \setminus Z) \xrightarrow{\sim} C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)$ constructed in the previous step. The second isomorphism is $\beta_*$, obtained from the Five Lemma in the fourth step. The composite $\beta_* \circ \varphi_*$ is the map induced on homology by the inclusion $(X \setminus Z, A \setminus Z) \hookrightarrow (X, A)$: tracing through the definitions, a relative cycle $c \in C_n(X \setminus Z)$ with $d(c) \in C_{n-1}(A \setminus Z)$ is sent by $\varphi$ to $[c] \in C_n^{\mathcal{U}}(X)/C_n(A)$, and then by $\beta$ to $[c] \in C_n(X)/C_n(A) = C_n(X, A)$, which is the image of $c$ under the inclusion $C_n(X \setminus Z) \hookrightarrow C_n(X)$ followed by the quotient $C_n(X) \twoheadrightarrow C_n(X, A)$. Therefore \begin{align*} H_n(X \setminus Z, A \setminus Z) \cong H_n(X, A) \quad \text{for all } n, \end{align*} and the isomorphism is induced by the inclusion map. This completes the proof of the Excision Theorem. [guided] Let us trace the full logical chain. We started with the inclusion of pairs $(X \setminus Z, A \setminus Z) \hookrightarrow (X, A)$ and wanted to show it induces isomorphisms on relative homology. The proof proceeded in three stages: 1. **Algebraic setup:** We constructed a commutative diagram relating the $\mathcal{U}$-small chain complex to the full chain complex, with $C_\bullet(A)$ common to both rows. 2. **The Small Simplices Theorem** gave us $\alpha_*: H_n^{\mathcal{U}}(X) \xrightarrow{\sim} H_n(X)$. The **Five Lemma** then upgraded this to $\beta_*: H_n(C_\bullet^{\mathcal{U}}(X)/C_\bullet(A)) \xrightarrow{\sim} H_n(X, A)$. 3. **The algebraic identification** $C_\bullet^{\mathcal{U}}(X)/C_\bullet(A) \cong C_\bullet(X \setminus Z, A \setminus Z)$ converted the abstract quotient into a recognisable relative chain complex. The power of this argument is that all the hard topology is concentrated in the Small Simplices Theorem. Once that theorem is available, excision is a formal consequence of diagram chasing (the Five Lemma) and the elementary algebraic identification of the quotient. The same strategy, applied to a different short exact sequence (with $C_\bullet(A \cap B)$ replacing $C_\bullet(A)$), yields the Mayer-Vietoris sequence. [/guided] [/step]