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On a branch where $(\rho ,\varphi )$ are valid local coordinates, the inclusion map is $j(\rho ,\varphi )=(x(\rho ),\rho \cos \varphi ,\rho \sin \varphi )$, with $x^2=1-a^2{\rho }^2$. Its differential givesStrictly this is first a vector field along a map on the surface. The displayed formula extends it canonically to a smooth vector field on all of ${\mathbb{R}}^3$. Its integral curves of a vector field through $(x_0,y_0,z_0)$ obey $\dot{x}=0$, $\dot{y}=-z$, $\dot{z}=y$, henceThese are circles around the $x$-axis, traversed with angular speed one; points on the axis are stationary. Since $K(x^2+a^2(y^2+z^2))=0$, curves starting on the ellipsoid stay there. The local angular chart can expire while the ambient flow map and the geometric curve remain well defined.

Suppose $X=A\cup B$ and that the interiors of $A$ and $B$ cover $X$. The Mayer–Vietoris theorem gives the long exact sequence in homologywhere all four displayed maps except the connecting homomorphism $\partial$ are induced by the relevant inclusion maps.

To prove it, let $C_{\ast }^{A,B}(X)$ be the chain complex generated by those singular simplices whose images lie wholly in $A$ or wholly in $B$. The sequenceis a short exact sequence of chain complexes. Repeated barycentric subdivision makes every singular simplex small enough to lie in $A$ or $B$, and the subdivision operator is chain homotopic to the identity. The inclusion $C_{\ast }^{A,B}(X)\hookrightarrow C_{\ast }(X)$ therefore induces an isomorphism on every homology group. The long exact homology sequence of a short exact sequence of chain complexes now gives the displayed sequence. Explicitly, if an $n$-cycle $z=a+b$ with $a\in C_n(A)$ and $b\in C_n(B)$, then $\partial a=-\partial b$ lies in $C_{n-1}(A\cap B)$ and represents the connecting class $\partial [z]$.

Take $X=\{0,1,2,\dots \}$, the nonnegative integers, and $Y=\{1,2,3,\dots \}$, the positive integers. Choose the injectionsHere $g$ is an inclusion map, while $f$ reaches only the positive even integers. In particular, $f$ is not a bijection: it misses every positive odd integer. Simply using the inverse function of $g$ on its image misses input zero.

The remedy is to follow the alternating injections starting at the missing input zero:The subscripts distinguish an input in $X$ from an output in $Y$, even when their numerical values agree. The resulting subset of $X$ is $A=\{0,2,6,14,30,\dots \}$. Its sequence of members satisfies $a_0=0$ and $a_{k+1}=2a_k+2$, giving $a_k=2^{k+1}-2$ by mathematical induction. Define the function

The image $h(A)=\{2,6,14,30,\dots \}=A\setminus \{0\}$ covers all the positive members of $A$ exactly once. The other inputs cover $Y\setminus A$ exactly once by staying fixed. These output sets are disjoint and their union is $Y$, proving that $h$ is a bijection. In particular, $h(0)=2$, $h(2)=6$, $h(6)=14$, while $h(1)=1$ and $h(4)=4$.

For completeness, its inverse function isThe familiar bijection $n\mapsto n+1$ also works, but this example shows something more specific: the two chosen injections can repair the missing zero by moving only a sparse sequence of inputs. The repair never has to shift the entire set.