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Let $V$ be a finite-dimensional real inner-product space and let $F:V\to V$ be continuous. If $⟨F(x),x⟩<0$ on the sphere $\Vert x\Vert =R$, then $F$ has a zero in the open ball. Otherwise the radial projection of $F$ gives a map of the ball that contradicts the Brouwer fixed-point theorem.

No closed disc retracts continuously onto its boundary. This is equivalent to Brouwer fixed-point theorem by the standard ray construction and the antipodal map.

For a single resource, assume unit-capacity calls arrive as a Poisson process of rate $a$, holding times have independent exponential distributions with mean $h$, and a call is rejected when all $C$ units are occupied. Let $\nu =ah$. Its occupancy is a birth-death process on $\{0,\dots ,C\}$, with birth rate $a$ below capacity and death rate $k/h$ in state $k$. Detailed balance for a birth-death process givesBy Poisson arrivals see time averages, the Erlang loss formula isThis derivation assumes exponential holding times; the standard insensitivity result extends the same formula to independent holding times of the same mean, but is not needed here.

A loss network with fixed routing has finitely many resources $j$, capacities $C_j$, and call types $r$ with a fixed nonempty set of resources. Type $r$ has Poisson arrivals of rate $a_r$, independent holding times of mean $h_r$, and offered load ${\nu }_r=a_r{h}_r$. It is accepted only if every resource on its route has a free unit, and otherwise is lost. In the displayed equations use the unit-incidence matrix $A_{jr}=1$ for $j\in r$ and zero otherwise. The equations concern this unit-demand model, rather than arbitrary multi-unit requirements.

The Erlang fixed point approximation replaces the correlated resource availabilities by independent ones. A type-$r$ call survives blocking at resources other than $j$ with approximate probability ${\prod }_{i\in r\setminus \{j\}}(1-B_i)$. Treating this filtering as Poisson thinning gives the reduced offered loadApply the single-resource formula to obtain $B_j=E({\rho }_j,C_j)$. Independence and Poisson thinning here are approximations; no such approximation is claimed for the exact network's joint stationary distribution.

Assume first $C_j\ge 1$. The right side is a continuous map from $[0,1{]}^J$ into itself, so the Brouwer fixed-point theorem gives existence. To prove uniqueness, let $m_C(\rho )=\rho [1-E(\rho ,C)]$ be the carried load of an Erlang loss resource, also the mean of the truncated Poisson occupancy. Differentiation yieldsThus $E$ increases continuously from zero to one and $m_C$ increases from zero to $C$.

Write $x_j=-\log (1-B_j)\ge 0$, and define ${\rho }_C(x)=E(\cdot ,C{)}^{-1}(1-e^{-x})$ and ${\ell }_C(x)=m_C({\rho }_C(x))$, with ${\ell }_C(0)=0$. Each ${\ell }_C$ is continuous and strictly increasing. A fixed point satisfiesThese are precisely the stationary equations for the convex potential for the Erlang fixed pointThe first term is convex; the second is strictly convex because each ${\ell }_{C_j}$ strictly increases. Hence $F$ is strictly convex. Its gradient cannot vanish at two distinct points: the difference of gradients dotted with the difference of points is strictly positive, with a positive contribution from every changed coordinate in the integral term. This also covers boundary coordinates $x_j=0$, using the right derivative there. Every fixed point has $B_j<1$, since its offered load is finite, so the change of variables is valid. There is exactly one fixed point. If a capacity is zero, its blocking probability is one; delete it and all routes using it before applying the argument to the remaining positive capacities. Unused resources have blocking probability zero.

The topological degree measures an oriented net number of sheets of a map. Its simplest integer-valued setting is a map $f:M\to N$ between connected oriented closed manifolds of the same positive dimension $n$. The hypotheses matter: an integer sign count needs orientations; compactness prevents preimages from escaping; and the same dimension makes a regular fiber discrete. A fundamental class gives the definition for continuous maps:Here $H_n(M;\mathbb{Z})$ and $H_n(N;\mathbb{Z})$ are generated by their chosen fundamental classes. Reversing either orientation changes the sign of the degree of a map between oriented manifolds. Disconnected sources contribute a sum over components; disconnected targets require a degree for each target component rather than one universal integer.

For a smooth map, choose a regular value $y$. The Sard theorem guarantees plentiful choices. Each point of $f^{-1}(y)$ has an invertible derivative, so the inverse function theorem makes it an isolated point. The fiber is finite by compactness of $M$. An oriented coordinate chart defines $\mathrm{sgn}\det d{f}_x$, and the degree as a sum of local degrees becomesAn empty fiber contributes zero. The sign is independent of the chosen oriented charts. Thus the mapping degree counts sheets with signs, not simply the number of preimages: orientation-preserving and orientation-reversing sheets cancel.

A differential-form argument both proves that this count is independent of $y$ and relates it to the topological definition. Around a regular value, take a small neighborhood $V$ whose full preimage is a disjoint union of neighborhoods mapped diffeomorphically onto $V$. Compactness rules out additional preimages approaching $V$ from elsewhere. Choose a smooth top-degree form $\eta$ supported in $V$, with ${\int }_N\eta =1$. The change of variables formula givesAlthough $\eta$ need not be everywhere positive, it represents the same normalized top cohomology class as any normalized volume form. The top de Rham cohomology of a compact connected oriented manifold says that two top forms with equal integrals differ by $d\beta$. Their pullback integrals agree, because the Generalized Stokes theorem gives ${\int }_{M}d(f^{\ast }\beta )=0$. Therefore the degree by integration of a pullback volume form isfor every smooth top form $\eta$. Pairing de Rham cohomology with fundamental classes identifies this integer with the homological definition. This proof also shows that every regular value gives the same signed sum.

The homotopy invariance of mapping degree makes it stable under continuous deformation. Homologically this follows from homotopic maps inducing the same map on homology. For a smooth homotopy $F:M\times [0,1]\to N$, a direct proof uses $d\eta =0$ and the Generalized Stokes theorem on the cylinder to getwith the cylinder orientation chosen to give the indicated boundary signs. Smooth approximation extends the geometric description to continuous maps, while the fundamental class definition already applies without differentiability. The multiplicativity of mapping degree follows from functoriality and givesA constant map has degree zero when $n>0$. If the mapping degree is nonzero, the map is onto: a point outside its image is a regular value with empty preimage. An orientation-preserving diffeomorphism has degree $1$, and an orientation-reversing one has degree $-1$. An orientation-preserving $k$-sheeted covering map has degree $k$; the signs of sheets must be included for other orientation conventions.

Examples show both the strength and the limits of the invariant. On the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, including negative integers and zero; it is the winding number. On the sphere $S^n$, an ambient reflection has degree $-1$, and the antipodal map has degree $(-1{)}^{n+1}$, because the determinant of $-I$ on the ambient ${\mathbb{R}}^{n+1}$ is $(-1{)}^{n+1}$. On the torus, an integer matrix $A$ induces a map of degree $\det A$, obtained by pulling back the constant top form. However, degree does not classify general manifold maps: the identity and an integer shear on the two-torus both have degree one but induce different maps on $H_1$, so are not homotopic.

The mapping degree gives useful obstruction and existence arguments. There is no retraction of a closed ball onto its boundary sphere: such a retraction would make the degree-one identity of the boundary null-homotopic, since it would extend across a contractible ball, contradicting homotopy invariance and the degree-zero constant map. This is the degree obstruction underlying the Brouwer fixed-point theorem. Also an everywhere nonzero tangent vector field on an even-dimensional sphere would, after normalization, produce the homotopy $\cos (\pi t)x+\sin (\pi t)v(x)$ from the identity to the antipodal map. Orthogonality makes every value a unit vector, but the endpoint degrees are $1$ and $-1$, impossible. This proves the corresponding Hairy ball theorem obstruction.

There are appropriate extensions rather than an unrestricted integer degree for every pair of manifolds. For a proper map between connected oriented manifolds without boundary of equal dimension, the regular-value count remains finite and is invariant; compactly supported top forms or locally finite homology replace the compact fundamental-class description. Without orientations, degree modulo two counts preimages modulo two and uses mod-two fundamental classes. For manifolds with boundary one uses relative mapping degree for maps of pairs, keeping the boundary away from the chosen target value. Maps of unequal dimensions, or nonproper maps where preimages can escape to infinity, do not in general admit this same signed sheet-count invariant. The central principle is an integer signed count stable under homotopy, with its hypotheses explicitly preserved.

Let $P$ be the row payoff matrix and $Q=P^T$. For $x$ in the probability simplex $\Delta$, write $u(x)=x^{T}Px$ and define the symmetric Nash gain mapThe map is a continuous function, has nonnegative coordinates, and sums to one. The Brouwer fixed-point theorem states that every continuous self-map of a nonempty finite-dimensional compact convex set has a fixed point. Apply it to $T:\Delta \to \Delta$, and let $x=T(x)$.

Set $G={\sum }_i{g}_i(x)$. The fixed-point equation gives $g_i(x)=G{x}_i$. If $G>0$, every positive $x_i$ has strictly positive gain, so $(Px{)}_i-u(x)=g_i(x)=G{x}_i$. Buta contradiction. Thus $G=0$ and every pure payoff $(Px{)}_i$ is at most $u(x)$. Since their $x$-weighted average equals $u(x)$, every supported action attains that maximum. Consequently $x$ is a best response to itself.

The column player's payoff vector against $x$ is $Q^{T}x=Px$, so exactly the same inequalities establish its best response. ThereforeThis supplies the whole fixed-point construction and fixed-point-to-equilibrium argument, rather than assuming Nash's theorem as a black box.

Sperner's lemma says that a triangulated triangle whose vertices are labelled $1,2,3$, with each outer vertex having its own label and each boundary edge using only its endpoint labels, contains a small triangle with all three labels. In fact the number of such triangles is odd. Count incidences of small triangles with edges labelled $1,2$. Interior edges contribute twice. On the outer $1,2$ edge, the number of changes between these labels is odd; the other two outer edges contribute zero. A small triangle contributes one incidence if it has all three labels, two if it uses both $1,2$ but not $3$, and zero otherwise. The total parity proves the assertion.

The resulting no-retraction theorem is: there is no continuous retraction of a closed disc onto its boundary. Work first with a closed triangle $P$. If $r:P\to \partial P$ fixes the boundary, colour each vertex of increasingly fine triangulations by an index of a largest barycentric coordinate of $r(v)$. The largest coordinate is at least $1/3$; on the boundary the colouring satisfies Sperner's lemma. Choose a tricoloured small triangle in each triangulation. By compactness and shrinking mesh, a subsequence of their three vertices converges to one point $x$. Continuity forces all three barycentric coordinates of $r(x)$ to be at least $1/3$, so $r(x)$ is the triangle's centre, contradicting $r(x)\in \partial P$. A homeomorphism of a disc with $P$ proves the disc assertion.

Brouwer fixed-point theorem states that every continuous self-map $F$ of a closed disc has a fixed point. If it had none, the ray from $F(x)$ through $x$ would meet the boundary beyond $x$, continuously in $x$. That intersection defines a retraction, since a boundary point is already the outgoing intersection. This contradicts the no-retraction theorem.

Finally fix $y\in {\mathbb{R}}^2$ and define $F_y(x)=y+x-g(x)$. The displacement bound implies $\Vert F_y(x)-y\Vert \le K$, so $F_y$ maps the closed radius-$K$ disc centred at $y$ into itself. Brouwer fixed-point theorem gives $F_y(x)=x$, which is exactly $g(x)=y$. Since $y$ was arbitrary, $g$ is surjective.