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Suppose two binary operations $\ast$ and $\cdot$ on one set share an identity element $e$ and satisfy the interchange ruleThen the operations coincide, becauseThey are also commutative:These substitutions prove the Eckmann-Hilton argument without cancellation assumptions. In the fundamental group of a topological group, the two operations are loop concatenation and pointwise group multiplication. They share the constant loop as identity on homotopy classes, so the fundamental group is Abelian.

A functor $F:\mathcal{C}\to \mathcal{D}$ is full and faithful when every mapis a bijection.

The functor category has functors $\mathcal{C}\to \mathcal{D}$ as objects and natural transformations as morphisms.

If $K/k$ is a finitely generated field extension, write $K=\mathrm{Frac}(A)$ for a finitely generated integral $k$-algebra $A$. The projective closure of the affine variety $\mathrm{Spec}A$ is an irreducible projective variety with function field $K$.

For an irreducible variety $X$ over a field $k$,Consequently birational irreducible varieties have the same dimension.

If $R$ is a nonzero commutative ring and $R^m\cong R^n$, reduction modulo a maximal ideal produces isomorphic vector spaces of dimensions $m$ and $n$. Hence $m=n$.

For a nonzero commutative scalar ring, every two bases of a finite-rank free module have the same cardinality. This cardinality is the rank.

A module $F$ is free on a set $S$ exactly when every function from $S$ to any module $N$ extends uniquely to an $R$-module homomorphism $F\to N$.

A basis $S$ of an $R$-module $M$ is a subset such that every element of $M$ has a unique expression as a finite $R$-linear combination of elements of $S$.

For $A$ terms with phase ratio $q$, the amplitude factor is $(1-q^A)/(1-q)$, with limiting value $A$ when $q=1$.

For the four-point one-sided second-derivative formula at unit spacing, take the signed errorIt annihilates cubic polynomials. The Peano kernel theorem givesThis kernel is nonnegative. On $[-1,0]$, put $u=t+1$ to get $K=u-u^3/3\ge 0$. On $[0,1]$, the inequality $2-t\ge 2(1-t)$ makes the remaining difference positive. On $[1,2]$, only $(2-t{)}^3/6$ remains. ThereforeIndeed, $\int K=L[(x+1{)}^4/24]=11/12$. The same polynomial has fourth derivative one and attains equality, so the constant is sharp. On a grid with spacing $h$, rescaling gives the bound $(11/12)h^2\Vert f^{(4)}{\Vert }_{\infty }$ for this second-derivative approximation.

If ${\Delta }^{k+1}f=0$, then $f$ is an integer linear combination of $1,\left(\genfrac{}{}{0ex}{}{n}{1}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ whenever $f$ is integer-valued. Successive differences recover the coefficients.

A discrete antiderivative of $f$ is a function $F$ satisfying $\Delta F=f$. Pascal's identity gives

When the source function varies approximately linearly with optical depth, the emergent intensity obeys $I_{\nu }(0,\mu )\simeq S_{\nu }({\tau }_{\nu }=\mu )$.

The relation is an approximation within radiative transfer, not the entire transfer theory.

For current $J=aP-\kappa \nabla P/2$, a scalar potential exists when the one-form ${\kappa }^{-1}a$ is exact. Then $D=\kappa /2$, $a=-D\nabla V$, and $J=-PD\nabla (V+\log P)$.

For unit edge resistances, the energy of a flow iswhere every unoriented edge is counted once.

The strength of a flow, also called its flow value, is net outflow from $s$, equivalently net inflow to $t$:The notation $|f|$ denotes this signed value, rather than an absolute value.^[ref]

The equality follows by summing the divergence of a flow over all vertices: the internal vertices contribute zero by flow conservation, while every edge cancels between its two endpoints. Hence $\mathrm{div}f(s)=-\mathrm{div}f(t)$. For the running flow, $|f|=3+2=2+3=5$.

Net outflow matters even when the source has incoming edges. As a separate example, give $s\to a$, $a\to s$ and $s\to t$ flow network edge capacities $2,2,3$ and flow amounts $2,2,3$. At $a$, flow conservation reads $2=2$. The source sends $5$ units along outgoing edges but receives $2$ back, so the flow value is $5-2=3$, exactly the amount arriving at $t$. The two-edge network circulation through $a$ raises total edge usage without increasing net delivery.

The divergence of a flow at a vertex $v$ is net outward flow. For original nonnegative edge amounts and their signed net record, respectively,Positive divergence of a flow means net supply, negative divergence of a flow means net demand, and zero is flow conservation. This is the same sign convention as flow balance.

In the running example, the divergences of a flow at $(s,a,b,t)$ are $(5,0,0,-5)$. Adding the four divergences of a flow gives zero: every edge contributes positively at one endpoint and negatively at the other. In a finite flow network, this cancellation explains why net supply at $s$ equals net demand at $t$.

The fractional part of a real number isIt is the amount left after removing the lower enclosing integer, even for negative inputs. Thus $\{-2.7\}=-2.7-(-3)=0.3$. It is not the signed string of digits obtained by truncation toward zero.

Translation by an integer leaves the fractional part unchanged: $\{x+m\}=\{x\}$. It is therefore a periodic function with period one, increasing from zero toward one on each unit interval, then returning to zero at the next integer. In Euclidean division $N=kq+r$ with $0\le r<k$, $\{N/k\}=r/k$; for $102/20$, this is $2/20=0.1$.

The notation is useful for following repeated motion modulo an integer. In normality and equidistribution under integer multiplication, the values $\{K^{n}x\}$ shift the digits of a positional notation; the fractional part discards the leading integer portion without losing the remaining digits. Its braces must be distinguished by context from the braces denoting a set.

The continuity equation givesDuring expansion with $\rho +3P<0$, the product $\rho a^2$ grows and $|{\Omega }^{-1}-1|\propto (\rho a^2{)}^{-1}$ decreases.