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Let $X$ have contractible universal cover, let $K$ be a path-connected simplicial complex, and let $i:K^1\hookrightarrow K$. A map $f:|K^1|\to X$ extends to $|K|$ exactly when its induced fundamental-group map factors through $i_{\ast }$. The factorization kills every 2-simplex boundary; all higher-dimensional boundary maps lift to the contractible universal cover and are null-homotopic.

For a path-connected locally path-connected base, any two based universal covers lift to one another. The two composites and the respective identity maps are based lifts of the same covering maps, so uniqueness of lifts makes the composites identities. The lift is therefore a homeomorphism.

Rational homology satisfiesIt records the free rank of integral homology and discards torsion.

The determinant $U(n)\to S^1$ sends a unitary loop to a circle-valued loop. A nonzero determinant winding number proves that the original loop class has infinite order.

The Werner–Holevo channel is $T(\rho )=(\mathrm{Tr}(\rho )I-{\rho }^T)/(d-1)$. Its normalized Choi matrix is the maximally mixed state on the antisymmetric subspace.

The normal derivative of a differentiable function $u$ along a hypersurface with unit normal $N$ is the directional derivativeHigher normal derivatives at a point $x$ may be defined by differentiating the one-variable function $s\mapsto u(x+sN(x))$ repeatedly at $s=0$.

The union-closed sets conjecture states that every finite nontrivial union-closed family has an element contained in at least half of its members.

An operator $T$ is a unilateral shift exactly when it is an isometry, $(\mathrm{Im}T{)}^{\perp }$ is one-dimensional, and ${\bigcap }_{n\ge 1}\mathrm{Im}(T^n)=\{0\}$. A unit vector $e_1$ in the first orthogonal complement is wandering, and $e_n=T^{n-1}{e}_1$ is a complete orthonormal basis.

For a bounded linear operator $T:X\to X$ on a complex Banach space, the point spectrum is precisely its set of eigenvalues:Here $I$ is the identity operator. Thus membership requires an actual nonzero eigenvector, whereas the spectrum of a bounded operator only requires failure of a bounded inverse. Every eigenvalue belongs to the spectrum, but the inclusion can be strict. The multiplication operator discussed below has an entire interval in its spectrum and empty point spectrum.^[ref]

If two tests have the same size and each has strictly greater power than the other at some alternative, neither test is uniformly most powerful.

For a second-order elliptic operator $Lu=a^{ij}{D}_{ij}u+b^i{D}_{i}u+cu$ with $c\le 0$, the inequality $Lu\ge 0$ on a bounded domain impliesAn exponential barrier reduces the non-strict inequality to a contradiction at a positive interior maximum.

A second-order operator is strictly elliptic at a point when its symmetric principal coefficient matrix is positive definite there. Uniform ellipticity strengthens this pointwise condition by requiring one positive lower bound throughout the domain.

If ${\mathbb{C}}^{\ast }$ embeds holomorphically in a compact Riemann surface $R$, the embedding extends across zero and infinity to a degree-one holomorphic map from the Riemann sphere to $R$. Hence $R$ is conformally the sphere.

Every plane domain whose complement contains at least two points is uniformized by the unit disc. A plane universal cover would give a nonconstant entire function omitting two values, contrary to the Little Picard theorem.

The quotients of $\mathbb{C}$ by free properly discontinuous conformal actions are $\mathbb{C}$, ${\mathbb{C}}^{\ast }$, and the complex tori $\mathbb{C}/\Lambda$. The translation group has respectively rank zero, one, or two.

The sphere has the Möbius group $PG{L}_2(\mathbb{C})$, the plane has the affine maps $z\mapsto az+b$ with $a\ne 0$, and the disc has the maps $e^{i\theta }(z-a)/(1-\bar{a}z)$ with $|a|<1$.

Every nonempty simply connected proper open subset of the complex plane is conformally equivalent to the unit disc.

The wired uniform spanning forest of an infinite graph is the weak limit of uniform spanning trees on finite exhaustions whose exterior vertices are identified to one wired boundary vertex.

If $n\ge 2k$ and $\mathcal{A}\subseteq [n{]}^{(k)}$ is intersecting, then $|\mathcal{A}|\le \left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$.

A sequence $f_n:X\to Y$ converges locally uniformly to $f$ when every point has a neighbourhood on which the convergence is uniform.