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Cover the sphere by complements of its two poles. Each open set is contractible, and their intersection deformation retracts to the equatorial sphere. The Poincaré lemma and homotopy invariance of de Rham cohomology make the two neighboring groups in the Mayer--Vietoris sequence for de Rham cohomology zero, so its connecting map is an isomorphism. Its inverse takes a top-degree form to the difference of its local primitives on the equator, up to the connecting-map sign convention.

The permutation group of a three-element set has six elements: the identity, three transpositions, and two three-cycles. A three-cycle and any transposition generate it, with relations $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$. Its alternating group is the normal cyclic subgroup of order three. It is the unique nonabelian isomorphism type of a group of order six.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

A linear map $A:(V,{\omega }_V)\to (W,{\omega }_W)$ is symplectic when it pulls back the target symplectic form to the source form. Nondegenerate bilinear form makes such a map injective. In equal finite dimensions it is an isomorphism. The automorphisms form the symplectic group and preserve the symplectic orientation on the ambient space and on every symplectic subspace.

Let $n\ge 2$ be even and let $E\subseteq {\mathbb{F}}_2^n$ be the vector subspace of even-weight vectors. Its dot product is an alternating bilinear form, since $x\cdot x={\sum }_i{x}_i=0$ on $E$. The orthogonal complement of $E$ in ${\mathbb{F}}_2^n$ is spanned by the all-one vector $1$, which lies in $E$. Consequently the radical of a bilinear form on $E$ is $⟨1⟩$, andis a nondegenerate symplectic vector space of dimension $n-2$. The symmetric group acts by coordinate permutations, preserving this alternating bilinear form.

At $n=6$, this group action is faithful. A weight-two vector and its complement have weights two and four, so a permutation fixing its quotient class must fix that two-element subset itself. Fixing every two-element subset forces the identity. The resulting embedding $S_6\hookrightarrow S{p}_4(2)$ is an isomorphism, because the symplectic group over a finite field order formula gives $2^4(2^2-1)(2^4-1)=720=6!$.

For modules over a commutative ring, there is a natural isomorphismgiven by $f\mapsto (m\mapsto (n\mapsto f(m\otimes n)))$.

For a $G$-representation $W$ and an $H$-representation $V$,On group-algebra tensors the isomorphism sends $x\otimes (w\otimes v)$ to $xw\otimes (x\otimes v)$.

A topological invariant is a property or mathematical object preserved by homeomorphisms of topological spaces. Connectedness and compactness are examples. The fundamental group is an invariant up to group isomorphism; it is also preserved up to isomorphism by homotopy equivalence, a stronger invariance property which does not hold for every topological invariant.

A smooth embedded submanifold has a neighborhood diffeomorphic to a neighborhood of the zero section of its normal bundle, with the zero section identified with the submanifold. One may prescribe the normal-bundle splitting of the derivative along that section. Choose a Riemannian metric making the splitting orthogonal and use its normal exponential map: its derivative there is the required isomorphism, so the inverse function theorem supplies local diffeomorphisms. A sufficiently small, possibly varying, fibre radius makes these fit into a tubular neighborhood.

Let $G\times G$ act on $\mathbb{C}G$ by $(g,h)x=gx{h}^{-1}$. Its character is zero unless $g$ and $h$ are conjugate, in which case it equals $|C_G(g)|$. Character column orthogonality gives the same character forwhere the $V_i$ run through the irreducible complex representations. Maschke's theorem therefore gives an isomorphism of the two representations.

An integral intersection form on a finite-rank free abelian group $L$ is unimodular if its adjoint $x\mapsto \lambda (x,-)$ is an isomorphism to the integral dual. In an integral basis, this is equivalent to the pairing matrix having determinant $1$ or $-1$, not merely nonzero determinant. For example, projective lines generate a unimodular intersection form on the Complex projective plane, with matrix $(1)$ in its complex orientation.

Choose rational $B=TU$, a representative $\dot{w}$, and $x$ with $x^{-1}F(x)=\dot{w}$. Set $T^{\prime }=xT{x}^{-1}$ and $U^{\prime }=F(xU{x}^{-1})$. Then $T^{\prime }$ is rational, andThe map sends $h$ to $hxB$. Quotienting only by the finite group $T^{\prime F}$ leaves fibres isomorphic to the positive-dimensional unipotent intersection when that intersection is nontrivial. For $w=1$ in ${\mathrm{SL}}_2$, this is an affine-line fibre over each rational flag, so the missing quotient cannot be ignored in a variety isomorphism.

Two representations of a functor have unique mutually inverse morphisms carrying their specified universal elements to one another. Composing those morphisms preserves each universal element, so injectivity of the representing bijection makes the composites identities. Arbitrary isomorphisms of the underlying objects need not preserve the specified elements.

Here is a short proof that uses only the row space. For a vector subspace $W\subseteq {\mathbb{F}}^n$, let $d_j$ be the dimension of its projection map onto the first $j$ coordinates, with $d_0=0$. In any reduced row echelon form with row space $W$, $d_j$ counts the pivot columns among those first $j$ columns: later-pivot rows project to zero, while earlier-pivot rows remain linearly independent. Thus the positions where $d_j-d_{j-1}=1$ determine all pivot positions from $W$ alone.

If those positions are $p_1,\dots ,p_r$, the map sending $w\in W$ to $(w_{p_1},\dots ,w_{p_r})$ is an isomorphism $W\to {\mathbb{F}}^r$: the reduced nonzero rows map to the standard basis. Its $i$th row is therefore the unique vector of $W$ with pivot-coordinate values equal to the $i$th standard basis vector. Any other reduced row echelon form with that row space has identical nonzero rows in identical order. At a fixed matrix size, the remaining rows must be zero. Together with existence by Gaussian elimination, this proves uniqueness.

Let $\mu :R{\otimes }_{k}R\to R$ be multiplication and ${\Omega }_R^{\mathrm{nc}}=\ker \mu$, with the outer $R$-bimodule structure. Thenis a derivation and composition with $D$ gives a natural isomorphismThe inverse sends $d$ toThe condition ${\sum }_i{r}_i{s}_i=0$ makes this map left as well as right $R$-linear. Every element of ${\Omega }_R^{\mathrm{nc}}$ is a sum of terms $D(r)s$, which proves uniqueness.

A universal property characterizes an object by a prescribed family of morphisms and a unique factorization of every competing family. It determines that object up to a unique compatible isomorphism. Examples include the pushout in a category and the universal property of the tensor product of modules.

An unlabelled combinatorial structure is a combinatorial structure considered up to an appropriate isomorphism, so the names assigned to underlying elements do not distinguish objects. For graphs, this means a bijection of vertices preserving adjacency. A different drawing or vertex numbering can therefore represent the same unlabelled graph.

On three labelled vertices there are $2^{\left(\genfrac{}{}{0ex}{}{3}{2}\right)}=8$ simple graphs, but only four unlabelled types: the empty graph, one edge with an isolated vertex, the two-edge path, and the triangle. Simply dividing eight by $3!$ cannot work because these types have different numbers of automorphisms.

More generally, relabelling gives a group action of the symmetric group on labelled structures. Unlabelled structures are its orbits, and the orbit-stabilizer theorem explains how automorphisms affect orbit size. Burnside's lemma counts the orbits by averaging fixed-point counts.

A category is well-copowered if, for each object $C$, the epimorphisms $C\to Q$ have only a set of equivalence classes, where two are equivalent through an isomorphism of their codomains compatible with the quotient maps. Equivalently, its opposite category is well-powered.

The standard-tableau right-ideal decomposition of $\mathbb{C}{S}_n$, tensored over that algebra with $V^{\otimes n}$, gives this direct sum of general linear group modules. With $e_t=h_t/H_{\lambda }$, the map $e_t\mathbb{C}{S}_n\otimes T\to e_{t}T$ is an isomorphism. Individual summands generally need not be symmetric group submodules; instead each is a Schur module for the commuting action.