Mayer vietoris sequence in singular cohomology

The dual of $j$ sends $w\in D^n$ to its two restrictions $a(w)$, while the dual of $i$ sends $(u,v)$ to $b(u,v)=u\mathrm{\mid }-v\mathrm{\mid }$, by [F1]. The map $a$ is injective because $j$ is onto. If $b(u,v)=0$, their values agree on common simplices; assign the value from $u$ to a simplex in $U$, and otherwise that from $v$. This gives a unique cochain on the small basis and its restrictions are $(u,v)$. Thus $\ker b=\mathrm{im}a$. Finally for a cochain $c$ on the intersection, extend it by zero to $U$ and pair this extension with zero on $V$; this maps to $c$. All maps commute with differential by [F1] and [F3], proving $0\to D\stackrel{a}{\to }E\stackrel{b}{\to }F\to 0$ termwise exact, including negative zero degrees.

Precomposition $I^{\mathrm{♯}}:C^{\ast }(X;G)\to D$ is a cochain homotopy equivalence. Indeed let $R$ be the chain homotopy inverse in [F2]. Precomposing with each of $R,I$ gives cochain maps in reverse order. If a chain homotopy satisfies $1-IR=\mathrm{\partial }T+T\mathrm{\partial }$, precomposition by $T_{n-1}$ gives $1-R^{\mathrm{♯}}{I}^{\mathrm{♯}}=\delta K+K\delta$; the other inverse homotopy gives the other composite. At degree zero the homotopy with index $-1$ is zero. Thus the induced map $\theta :H^n(X;G)\to H^n(D)$ is an isomorphism. No Hom exactness theorem is used.

For a cocycle $c\in F^n$, lift it to $e\in E^n$ by step 1.1. Then $b\delta e=\delta c=0$, so $\delta e=a(d)$ for a unique $d\in D^{n+1}$. Injectivity of $a$ and ${\delta }^2=0$ imply $\delta d=0$. If $e$ changes by $a(t)$, then $d$ changes by $\delta t$. If $c$ changes by $\delta c_0$, lift $c_0$ to $e_0$ and replace $e$ by $e+\delta e_0$, which leaves $d$ unchanged. Hence ${\mathrm{\partial }}_D[c]=[d]$ is well-defined; lifts of sums show it is additive. Define the asserted connector by $\mathrm{\partial }={\theta }^{-1}{\mathrm{\partial }}_D$. This uses the same positive sign as [F5]. Concretely, one may lift by $(\tilde{c},0)$ where $\tilde{c}$ is zero extension to $U$; the small cocycle $d$ restricts to $\delta \tilde{c}$ on $U$ and zero on $V$.

At $H^n(E)$, images from $D$ map to zero. If a cocycle $e$ has $be=\delta c_0$, lift $c_0$ to $e_0$ and replace $e$ by $e-\delta e_0$. This is a cocycle in $\ker b$, hence equals $a(d)$ for a unique cocycle $d$, since $a$ is injective and commutes with differential. Thus this kernel equals the image from $H^n(D)$. The identification $H^n(E)=H^n(U;G)\oplus H^n(V;G)$ is coordinatewise: cycles and boundaries are respectively pairs of cycles and pairs of boundaries.

At $H^n(F)$, a class represented by $be$ with $e$ closed has zero connector. Conversely, if the connecting class of $c$ vanishes, its lift satisfies $\delta e=a(\delta t)$ for some $t\in D^n$. Then $e-a(t)$ is a cocycle lifting $c$. At $H^{n+1}(D)$, a connecting representative $d$ satisfies $a(d)=\delta e$, so its image is zero. Conversely if a cocycle $d$ has $a(d)=\delta e$, then $be$ is a cocycle and its connector is $[d]$. These arguments establish exactness at both positions in both directions. They include the initial $H^0(D)$ injection because a global primitive in $E^{-1}$ is necessarily zero.

Replace $H^{\ast }(D)$ by $H^{\ast }(X;G)$ through $\theta$. The map to the two open sets is the actual pair of restrictions because $a{I}^{\mathrm{♯}}$ restricts absolute cochains. The next map is the stated difference by step 1.1. Steps 2.2 and 3.1 give the exact sequence with connector from step 2.1. A map of ordered covers sends small chains to small chains and commutes with their inclusions and $i,j$; hence its cochain maps commute with $a,b$ and the canonical $I^{\mathrm{♯}}$. Sending a lift $e$ through these maps gives a lift of the image of $c$, and its differential is the image of $a(d)$. Thus it commutes with ${\mathrm{\partial }}_D$, and the commuting isomorphisms $\theta$ imply commutation with $\mathrm{\partial }$. Coefficient postcomposition has the same property. No naturality of the chosen chain inverse $R$ or of zero extension is required.

This proves the statement, naturally with its printed sign. If $U$ or $V$ is empty, the sequence has the remaining restriction as an identity and zero intersection groups. If $U=V=X$, then $\alpha$ is diagonal, $\beta (u,v)=u-v$, and the connector is zero since $c\mapsto (c,0)$ is a cochain lift. These include point spaces and empty $X$; zero coefficients and negative degrees give zero terms. At degree zero there are no negative primitives, as checked in step 3.1. The two-coordinate constructions need no infinite choice; the small-chain construction uses least subdivision depths and all cochain extensions use the specified zero. No AC is used.