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Assign each arriving car independently to a uniformly chosen lot. The Poisson thinning theorem gives three independent Poisson processes, each of rate $\lambda /3$. For an interval of length $t$, their joint probability generating function isThe factorization proves independence of the counts in that interval; independence over disjoint intervals proves independence of the processes.

For a renewal gap $T$ of mean $m$, the equilibrium first-delay density is $\mathbb{P}(T>t)/m$. Here $m=3/\lambda$ and the Erlang distribution survivor is $e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. The equilibrium delay of an Erlang renewal process is thereforeThis is an equal mixture of shapes one, two and three. For each equilibrium stream use this first delay, followed by independent shape-three gaps. A joint stationary version of the cyclic routing is obtained by choosing the initial position in the three-label cycle uniformly at random, independently of the Poisson arrivals. Its marginal streams have the stated equilibrium law, but remain dependent on one another.

Streams $a$ and $b$ are delayed renewal processes; $c$ is an ordinary renewal process. Their first delays have Erlang shapes one and two, while subsequent gaps have shape three. Stream $c$ has shape three for its first gap as well.

Within each stream every subsequent gap is the sum of three consecutive original exponential random variables, hence has an Erlang distribution of shape three and rate $\lambda$. Successive gaps within that stream use disjoint original gaps, so they are independent and identically distributed, and independent of its first delay. Gaps from different streams can overlap in their constituent original gaps, so the three streams are not independent.

Write the successive gaps of the rate-$\lambda$ Poisson process as independent exponential random variables $E_j$. The first arrivals in the three streams occur at $E_1$, $E_1+E_2$, and $E_1+E_2+E_3$. They have Erlang distributions of shapes one, two and three:

The uniqueness theorem for measures says that two measures agreeing on a generating pi-system agree on its generated sigma-algebra, provided they are finite with the same total mass, or there is a common increasing exhaustion by pi-system sets of finite measure.

For the finite case let $\mathcal{D}=\{A:\mu (A)=\nu (A)\}$. Equality of total masses gives closure under complements, and countable additivity gives closure under disjoint countable unions. Thus $\mathcal{D}$ is a Dynkin system containing the generating pi-system. The pi-lambda theorem puts the entire generated sigma-algebra inside $\mathcal{D}$.

For the exhaustion version, let $E_n\uparrow E$ with $E_n$ in the pi-system and $\mu (E_n)=\nu (E_n)<\infty$. The finite measures $A\mapsto \mu (A\cap E_n)$ and $A\mapsto \nu (A\cap E_n)$ agree on the generating pi-system, since it is intersection-closed. The finite proof applies, and continuity from below of a measure as $n\to \infty$ gives $\mu (A)=\nu (A)$ for every measurable $A$. A finiteness hypothesis is essential to this standard extension-uniqueness statement.

${\mathcal{E}}_1$ is a Dynkin system but not a pi-system. The whole ten-point set has even cardinality, complements preserve parity, and disjoint unions of even sets are even. Two even sets can intersect in a singleton, for example $\{1,2\}$ and $\{2,3\}$.

${\mathcal{E}}_2$ is neither. The infinite set $\mathbb{N}\setminus \{1\}$ belongs to it, but its singleton complement does not, so it is not a Dynkin system. The two infinite sets $\{1\}\cup \{2,4,6,\dots \}$ and $\{1,3,5,\dots \}$ intersect in the excluded singleton, so it is not a pi-system.

${\mathcal{E}}_3$ is a pi-system but not a Dynkin system. Intersections of bounded open intervals are bounded open intervals or empty. However, the family does not contain the whole underlying set $\mathbb{R}$. None of the three families is a sigma-algebra.

Every sigma-algebra is both a pi-system and a Dynkin system. Conversely, in a family having both properties, intersections and complements give finite unions and differences. For any sequence $(A_n)$ in it, replace $A_n$ by the disjoint sets $B_n=A_n\setminus {\bigcup }_{j<n}{A}_j$, which still belong to the family. Their disjoint union equals ${\bigcup }_n{A}_n$, proving closure under arbitrary countable unions. Thus the family is a sigma-algebra.

A pi-system is a family closed under finite intersections. A Dynkin system contains the whole underlying set, is closed under complements in it, and is closed under countable pairwise-disjoint unions.

A submersion from a nonempty compact manifold to ${\mathbb{R}}^{\ell }$, $\ell \ge 1$, would be surjective by the preceding result, since the target is connected. Its image would also be compact, contradicting noncompactness of ${\mathbb{R}}^{\ell }$. Thus no such submersion exists. The nonempty-domain qualification excludes the vacuous empty-manifold case.

Coordinate projections are open maps, so the local form of the submersion proves that $f$ is open. If $X$ is compact and nonempty, its continuous image is compact and hence closed in the Hausdorff manifold $Y$. Therefore $f(X)$ is a nonempty set both open and closed in connected $Y$, giving $f(X)=Y$.

At a point where $df$ is surjective, choose charts and reorder the domain coordinates so that an $\ell \times \ell$ minor of the derivative is invertible. The map $F=(f_1,\dots ,f_{\ell },x_{\ell +1},\dots ,x_k)$ has invertible derivative. By the inverse function theorem it gives local coordinates, in which $f$ is the projection $(u_1,\dots ,u_k)\mapsto (u_1,\dots ,u_{\ell })$. This is the submersion theorem.

The genus-two surface has fundamental group $G=⟨a_1,b_1,a_2,b_2\mid [a_1,b_1][a_2,b_2]=1⟩$. Connected double covering spaces correspond to index-two subgroups. Each is the kernel of a nonzero homomorphism $G\to C_2$; conjugacy creates no identifications because these kernels are normal. The commutator relation is automatic in $C_2$, so there are $2^4-1$ choices:To obtain a surjection onto the free group $F_2=⟨x,y⟩$, send $a_1\mapsto x$, $a_2\mapsto y$ and $b_1,b_2\mapsto 1$. The defining relation again vanishes, and the images generate $F_2$.

The polygon presentation of the Klein bottle, or Seifert-van Kampen theorem applied to its cell structure, givesSend $a$ to $(12)$ and $b$ to $(123)$ in $S_3$. Conjugation by $(12)$ inverts $(123)$, so this defines a surjective group homomorphism. Since $S_3$ is nonabelian, the fundamental group is nonabelian.

Modulo five, $X^2+26=(X-2)(X+2)$. Choose ${\mathfrak{p}}_5=(5,2+\alpha )$; then $(5)={\mathfrak{p}}_5{\overline{\mathfrak{p}}}_5$, and $N(2+\alpha )=4+26=30$. The element $2+\alpha$ belongs to ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\mathfrak{p}}_5$. Their product has norm $2\cdot 3\cdot 5=30$, soThe imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{104}<7$. Every ideal class has an integral representative of norm at most six, whose prime ideal factors lie over two, three or five. Thus these prime ideals generate the ideal class group. Conjugate classes are inverses, and the displayed relation eliminates the class over five. The remaining generators have orders two and three, so their product has order six and generates the whole group:

The ideal ${\mathfrak{p}}_2=(2,\alpha )$ has ideal norm two. Its square is contained in $(2)$ and has norm four, so $(2)={\mathfrak{p}}_2^2$. A generator would have field norm $u^2+26v^2=2$, impossible for integers $u,v$. Thus its ideal class has order two.

Modulo three, $X^2+26=(X-1)(X+1)$, giving $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ with ${\mathfrak{p}}_3=(3,1-\alpha )$. A generator of ${\mathfrak{p}}_3$ would have norm three, again impossible. If ${\mathfrak{p}}_3^2$ were principal, its generator would have norm nine. The only possibilities are $\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Finally $N(1-\alpha )=27$. This element lies in ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$, because $\alpha \equiv -1$ in the latter quotient. Unique unique factorization of ideals in a number field and norms therefore give

Write $\alpha =\sqrt{-26}$. Since the squarefree radicand is not congruent to one modulo four, the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$. The discriminant of the basis $(1,\alpha )$ is $4(-26)$, giving

Because $b$ is not an integer, the coefficient $\sin (\pi b)$ is nonzero. The expression of $B$ in terms of the entire function $J$ can therefore have poles only at integral $z$, and at most simple ones. Positive integers are removable: $B(z,b)$ is already analytic there for $\mathrm{Re}b>0$, and analytic continuation in $b$ givesThus $J$ vanishes at those integers, cancelling the denominator zero. The Gamma function formula $B(z,b)=\Gamma (z)\Gamma (b)/\Gamma (z+b)$ identifies the remaining singularities:For nonintegral $b$, these residues are nonzero, so every listed singularity is a genuine simple pole.

Contract the loops in the Pochhammer contour onto the interval $(0,1)$. When $\mathrm{Re}z,\mathrm{Re}b>0$, the small-circle contributions tend to zero. The two clockwise circuits change the branches by $e^{-2\pi ib}$ and $e^{-2\pi iz}$, and the return circuits undo them. The four interval contributions sum toUsing the Beta function and $1-e^{-2\pi iu}=2i{e}^{-\pi iu}\sin (\pi u)$ gives

$J$ is entire in $z$. Along the fixed Pochhammer contour, the continued logarithms of $t$ and $1-t$ are bounded, and the contour stays away from both branch points. The integrand $\exp ((z-1)\log t+(b-1)\log (1-t))$ is entire in $z$ with uniform bounds on each compact parameter set. Differentiation under the contour integral is therefore valid to every order.