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Suppose $f$ is continuously differentiable near $a$ with $f^{\prime }(a)\ne 0$. It has a local inverse function $g$ near $b=f(a)$, andThe chain rule applied to $g(f(x))=x$ gives the formula. The nonzero derivative condition also ensures local invertibility; it does not assert a global inverse function on an arbitrary larger domain of a function. For $f(x)=x^2$ at $a=2$, the corresponding output is $b=4$, and the square root's derivative there is $1/4$. At $a=0$, this division is unavailable: the nonnegative restriction still has a square root inverse, but its finite derivative at zero does not exist. ^[ref].

For a real-valued function drawn in the plane, the horizontal line test asks whether every horizontal line intersects its function graph at most once. This is exactly injectivity, since two intersections at the same height mean two inputs have the same output. A missing height is compatible with injectivity, so an inverse function is then defined on the image of a function; it is an inverse on the entire specified codomain only when surjectivity also holds.

A quasiperiodic orbit on a torus has angular coordinates $\theta (t)=\theta (0)+\omega t$ modulo $2\pi$. It is periodic exactly when the nonzero frequencies are commensurate; rationally independent frequencies produce a dense orbit.

Replace a degree-$d$ binary form by the formal power $(a_{0}x+a_{1}y{)}^d$ and recover its coefficients with the linear rule $a_0^{d-i}{a}_1^i\mapsto f_i$. For a coefficient invariant of degree $r$, use $r$ symbolic letters, each of degree $d$, polarize and then apply this rule to each letter. Products of brackets $(ab)=a_0{b}_1-a_1{b}_0$ generate the invariant expressions before the final coefficient substitution.

Geometric invariant theory constructs algebraic quotients of group actions by a reductive algebraic group using invariant sections and a chosen linearization of a line bundle. An affine quotient is the spectrum of the polynomial invariant ring; a projective quotient uses homogeneous invariants and excludes the points where all positive-degree invariants vanish. Stability separates the orbits which give well-behaved geometric quotient points.

For a linear group representation, the common fixed vectors of a subgroup form an invariant subspace for any equivariant map restricted to it. Its dimension is central to symmetry-based branching theorems. It need not be invariant under the entire group; conjugate subgroups have conjugate fixed-point subspaces.

Using the double-negation Boolean algebra, form well-founded names whose immediate subnames have truth-value weights. Define valued membership and equality by recursion on the underlying well-founded trees. Logical conjunction and universal quantification use meets; disjunction and existence use Boolean joins. Explicit names for pairs, unions and subsets verify the elementary set axioms, and collection for Boolean-valued names handles unbounded witnesses. The resulting internal valued universe validates classical ZF. This is a negative inner interpretation, not an assertion that the external intuitionistic universe satisfies excluded middle.

Quenched intrinsic curvature of an elastic filament is fixed during a mechanical experiment, although different filaments can have different random intrinsic shapes. Ensemble averaging is then over these shapes after mechanical minimization. It differs from averaging over time-dependent thermal bending fluctuations in a worm-like chain.

For a planar elastic filament under positive constant extension force $F$, the quadratic force-controlled energy isThe higher-order Euler-Lagrange equation is $B{\zeta }^{\prime \prime \prime \prime }-F{\zeta }^{\prime \prime }=B{\kappa }_0^{\prime \prime }$. On a periodic interval or in a bulk Fourier transform treatment, every nonzero mode obeys the displayed transfer relation. The length deficit is $\frac{1}{2}\int ({\zeta }^{\prime }{)}^{2}dx$, so its ensemble value is determined by the power spectrum of the quenched intrinsic curvature. Boundary conditions must be specified for a finite filament.

Moment-free ends require ${\zeta }^{\prime \prime }={\kappa }_0$ there, so a highly stretched finite filament can retain curvature in boundary layers of width $\ell =\sqrt{B/F}$. For constant intrinsic curvature $\kappa$, zero transverse end forces give slope ${\zeta }^{\prime }(x)=\kappa \ell \sinh [(x-L/2)/\ell ]/\cosh [L/(2\ell )]$. Direct integration yieldsIts large-force deficit is ${\kappa }^2{\ell }^3/2$, of order $F^{-3/2}$, even though the smooth bulk curvature derivative vanishes. This shows why endpoint assumptions matter in force-extension of an intrinsically curved filament.

For spherical thermal hydrostatic equilibrium, $dP/dr=-{\rho }_{g}GM(<r)/r^2$. Inserting the ideal gas law $P={\rho }_g{k}_{B}T/(\mu m_p)$ with constant mean molecular weight gives the displayed estimator. The density ${\rho }_g$ is gas density, while $M$ includes every gravitating component. A varying $\mu$ adds $-d\log \mu /d\log r$ inside the logarithmic bracket. Nonthermal pressure requires differentiating the total supporting pressure rather than thermal pressure alone.

A sequence of nested intervals keeps every new enclosure inside the previous one:Each step adds a restriction without discarding an earlier restriction. This is useful when an exact real number is described by increasingly precise bounds rather than by a finite decimal expansion. The bisection method, for instance, produces nested intervals that keep enclosing a zero of a function.

For a simple example, the closed intervals $I_n=[1-1/n,1+1/n]$ are nested and all contain $1$. Their lengths $2/n$ tend to zero, so the nested interval theorem leaves exactly that point. The family $J_n=[1-1/n,2+1/n]$ is also nested, but its intersection is the whole closed interval $[1,2]$. Nesting alone does not force uniqueness.

The open intervals $K_n=(0,1/n)$ illustrate a different possibility. Every finite collection has a nonempty intersection, but the intersection of all of them is empty: any positive candidate eventually exceeds $1/n$, and zero is excluded from every member. Each finite row in the diagram contains points; the full infinite family need not.

For a continuous function from an interval to itself with a three-cycle $a<b<c$, reflection if needed gives $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b],J=[b,c]$. The intermediate value theorem gives $F(I)\supseteq J$ and $F(J)\supseteq I\cup J$. Successive closed-interval pullbacks realize every cyclic covering word by a point fixed under the corresponding iterate. The word $I{J}^{n-1}$ has a unique visit to the interior of $I$ per cycle, forcing least period $n$; the only boundary ambiguity is the original three-cycle, which realizes $n=3$ and cannot realize any other such word. A covering of $J$ by itself also gives a fixed point. Thus every positive least period occurs.

The interval square range treats squaring as one function of one input. For $X=[a,b]$ it is exactlyIt retains the relationship that ordinary interval multiplication loses: for $Z=[-1/2,1/2]$, the interval square range is $[0,1/4]$, whereas the independent-input product $ZZ$ is $[-1/4,1/4]$.

Consequently, completing the square for $x(1-x)$ and using the interval square range givesThis is the exact image of a function on $X=[0,1]$. Merely rewriting the expression and evaluating $[1/4,1/4]-ZZ$ gives $[0,1/2]$, so it does not achieve the same sharpening. A mathematical $Z^2$ must be interpreted explicitly here: a square range operation and an independent-input interval product have different meanings. Choosing an operation that preserves known structure can save many steps of interval subdivision.

A natural interval extension evaluates a chosen expression by replacing its arithmetic operations with interval arithmetic operations, its constants with enclosing closed intervals, and any additional functions with valid interval enclosures. Each intermediate result must remain defined on its inputs. The resulting $F(X)$ encloses the actual image of a function $f(X)$, but its width depends on the expression used. Rump's INTLAB tutorial describes this inclusion principle and warns that repeated inputs can produce severe overestimation.^[ref]

For a proper morphism of varieties $f:X\to Y$, a Cartier divisor $D$ on $Y$, and an algebraic cycle $\alpha$ on $X$, pullback of $D$, intersection with $\alpha$, and proper pushforward satisfy the displayed identity in the Chow group. In particular, for a generically finite morphism of surfaces of degree $r$, $(f^{\ast }D{)}^2=r{D}^2$. For a curve $C$ contracted to a point, $f_{\ast }C=0$, so $f^{\ast }D\cdot C=0$.

For a proper integral Cartier curve $C$ on a smooth algebraic surface, its self-intersection is $\mathrm{deg}{\mathcal{O}}_C(C)$. If $C$ is smooth, this is the degree of its normal line bundle. It is defined over any algebraically closed field, without requiring an associated real or complex manifold.

// Target: geometry-and-topology.bigb

For $\dim L=2n-1$ and lower middle perversity, the suspension with two distinct vertices has $I{H}_i(\mathrm{Susp}L)=I{H}_i(L)$ for $i<n$, zero for $i=n$, and $I{H}_{i-1}(L)$ for $i>n$. Cover it by its two open cones. The Mayer–Vietoris sequence compares their truncated groups with $I{H}_{\ast }(L)$; below $n$ the overlap map is the injective signed diagonal, and above $n$ both cone groups vanish.

Here $H$ is the hyperbolic form with matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The degree-two cohomology consists of the two factor orientation classes and $4gh$ products of degree-one classes. The factor classes pair once and have zero squares. In symplectic bases $a_i,b_i$ and $c_j,d_j$, $(a_i{c}_j)(b_k{d}_l)=-{\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$ and $(a_i{d}_j)(b_k{c}_l)={\delta }_{ik}{\delta }_{jl}{\omega }_1{\omega }_2$. Each pair $(i,j)$ contributes two hyperbolic forms; the fiber classes contribute one more. The sign in the first formula comes from interchanging two odd-degree factors.

For a closed oriented smooth four-manifold, $b_2^{+}$ is the dimension of a maximal positive-definite subspace of its real intersection form. A symplectic form forces $b_2^{+}\ge 1$, because its cohomology class has positive square. Reversing orientation exchanges the positive and negative indices.