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For $G:\mathcal{D}\to \mathcal{C}$, a functor-split coequalizer pair is a parallel pair $f,g:Y\rightrightarrows Z$ such that its image admits a split coequalizer in $\mathcal{C}$. In one orientation this means morphisms $q:GZ\to Q$, $s:Q\to GZ$ and $t:GZ\to GY$ satisfyThese equations prove that $q$ is a coequalizer: if $hGf=hGg$, then $h=hGft=hGgt=hsq$, so $hs$ factors $h$; uniqueness follows from $qs=1$. They remain true under every functor, making a split coequalizer an absolute colimit.

To say that $G$ reflects such coequalizers means that any fork $Y\rightrightarrows Z\stackrel{p}{\to }W$ in $\mathcal{D}$ whose image is a split coequalizer is itself a coequalizer in $\mathcal{D}$. This is a reflection assertion about an existing fork; it does not by itself require lifting every split fork from $\mathcal{C}$, or creating its splitting maps in $\mathcal{D}$.

For $(A,a)$, consider the reflexive free-algebra presentation of a monad algebraBoth parallel morphisms are monad algebra morphisms: ${\mu }_A$ satisfies the monad associativity law, while $Ta$ is the image under $F^T$ of $a$. The map $a$ is a monad algebra morphism because $a{\mu }_A=aTa$, and the same equation says it equalizes the pair.

Suppose $h:F^T(A)\to (B,b)$ is a monad algebra morphism with $h{\mu }_A=hTa$. Put $k=h{\eta }_A:A\to B$. Naturality of $\eta$ and the unit law giveMoreover, since $h$ is a monad algebra morphism,Thus $k$ is a monad algebra morphism and factors $h$ through $a$. If $k^{\prime }a=h$, then $k^{\prime }=k^{\prime }a{\eta }_A=h{\eta }_A=k$, proving uniqueness. Therefore every monad algebra is the stated coequalizer of free algebras.

The underlying fork is a split coequalizer: take $q=a$, $s={\eta }_A$ and $t={\eta }_{TA}$. Then $qs=1_A$, ${\mu }_{A}t=1_{TA}$, and $(Ta)t=sa$. Its splitting maps need not be monad algebra morphisms; the preceding argument separately proves the coequalizer in ${\mathcal{C}}^T$.

Define the free algebra functor $F^T$ by $F^{T}X=(TX,{\mu }_X)$ and $F^{T}f=Tf$. The monad laws make ${\mu }_X$ an algebra action, and naturality of $\mu$ makes $Tf$ a monad algebra morphism. For an algebra for a monad $(A,a)$, definewith proposed inverse $g\mapsto aTg$. This inverse is a monad algebra morphism, sinceOne composite is $aTg{\eta }_X=a{\eta }_{A}g=g$. For a monad algebra morphism $h$, the other is $aT(h{\eta }_X)=aThT{\eta }_X=h{\mu }_{X}T{\eta }_X=h$. Both formulas commute with composition in a category in $X$ and with monad algebra morphisms in $(A,a)$, so the bijection is natural. Thus

A monad on $\mathcal{C}$ consists of an endofunctor $T$ and natural transformations $\eta :1\to T$, $\mu :T^2\to T$ satisfying, at every object $A$,These are the associativity and unit laws. An algebra for a monad is $(A,a:TA\to A)$ withA monad algebra morphism $h:(A,a)\to (B,b)$ is a morphism $h:A\to B$ satisfying $ha=bTh$. Identity morphisms satisfy this equation, and if $h,k$ satisfy it, then $kha=kbTh=cTkTh=cT(kh)$. Hence these objects and morphisms form the Eilenberg-Moore category ${\mathcal{C}}^T$, with the forgetful functor $U(A,a)=A$ and $Uh=h$.

Define $\mathcal{T}$ to be the family of all unions of members of $\mathcal{B}$, including the empty union. We will verify that this is a topology and that $\mathcal{B}$ is a basis of a topology. The two conditions play different roles: the covering condition supplies the whole space, while the local intersection condition supplies closure under finite intersections.

Use the least-upper-bound property as the least upper bound axiom: every nonempty set of real numbers bounded above has a least upper bound in $\mathbb{R}$. It implies convergence of a bounded increasing sequence: if $L$ is the supremum of its terms, then for every $\epsilon >0$ some term exceeds $L-\epsilon$, and all subsequent terms lie between that term and $L$. A bounded decreasing sequence converges by applying this argument to its negative.

The alternating series test states that if $b_n\ge 0$, $b_{n+1}\le b_n$, and $b_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n-1}{b}_n$ converges. To prove it, let $S_N$ denote its partial sums. The even partial sums are increasing because $S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\ge 0$. The odd partial sums are decreasing because $S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\le 0$. Moreover,Completeness of the real numbers therefore gives a limit of a sequence for each subsequence, denoted $L_e,L_o$ respectively for the even and odd subsequences. Their difference is $S_{2m+1}-S_{2m}=b_{2m+1}\to 0$, so $L_e=L_o$. Every sufficiently late partial sum belongs to one of these two subsequences, proving convergence of the full sequence of partial sums. Multiplying the terms by $-1$ leaves convergence unchanged.

Let $f$ be continuous on a closed bounded interval $[a,b]$. First prove that $f$ is a bounded function. If it failed, we could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x_{\ast }$. Because the interval is closed, $x_{\ast }\in [a,b]$. Continuity gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j\to \infty$.

Now let $M=\sup \{f(x):x\in [a,b]\}$, which exists by completeness of the real numbers and the bound just proved. Choose $y_n\in [a,b]$ with $f(y_n)>M-1/n$. A convergent subsequence has limit of a sequence $y_{\ast }\in [a,b]$, and continuity gives $f(y_{\ast })=M$. Applying the same argument to $-f$ gives attainment of the infimum. This proves the extreme value theorem: the maximum and minimum both exist and are attained. The argument also covers a one-point interval, where the conclusions are immediate.

Let $N$ be the $t\times b$ incidence matrix of a set system. The balanced incomplete block design overlap counts give $N{N}^T=(r-\lambda )I+\lambda J$. On the zero-sum subspace its eigenvalue is $r-\lambda =r(t-k)/(t-1)>0$; on the constant vector its eigenvalue is $r+(t-1)\lambda >0$. Therefore $N{N}^T$ has matrix rank $t$. Since a product cannot have larger rank than either factor,This proves Fisher's inequality for block designs directly. The incomplete, nontrivial hypotheses ensure the positive eigenvalue needed for this argument.

The identities require $r=k$ and $k(k-1)=2\cdot 3=6$, so $k=r=3$. Take treatments $1,2,3,4$ and the four blocksEach treatment is absent from exactly one block, hence occurs three times. Each pair is absent from the two blocks missing one of its members, hence occurs together in the other two. This is the required symmetric balanced incomplete block design; here $r-\lambda =1$ is indeed a square.

Symmetry forces $r=k$, and the pair-counting identity becomes $k(k-1)=2\cdot 45=90$. Its only positive integral solution is $k=r=10$. Since the number of blocks is even, the even-order symmetric design square obstruction requires $r-\lambda =8$ to be a square. It is not. No design with these parameters exists.

The transformed values are Cox–Snell residuals. Under a correct common survival function and a good fitted cumulative hazard function, the complete transformed failure times should have unit exponential distribution. Transform the observed times and retain their original censoring indicators, then estimate residual survival with a Kaplan–Meier estimator. Its target isso a plot of its logarithm against $u$ should lie approximately on a line of slope $-1$. Censored residuals must not be treated as fully observed failure times.

There is a useful caution for this particular common nonparametric fit. The shared-fit Cox–Snell residuals reproduce the fitted survival curve property means that the pooled residual plot is largely built into the transformation. With event ordering and compatible tie processing preserved, at a transformed failure time $u_j=\hat{H}(t_j)$ its residual Kaplan–Meier estimator equals $\hat{S}(t_j)=e^{-u_j}$. Finite steps, coalesced censoring times and estimation uncertainty still matter, and an infinite final residual arises if the fitted survivor becomes zero. Separate group plots are more informative about a common-distribution assumption than this pooled fit alone.

True. The preceding distributional argument alone shows that almost-sure convergence to zero cannot hold with probability one. To establish almost-sure nonconvergence, rather than merely failure of a probability-one convergence claim, use the Kolmogorov zero-one law.

Let $A=\{B_n/\sqrt{n}\to 0\}$. For every fixed $m$, subtracting $B_m/\sqrt{n}\to 0$ shows thatThus $A$ belongs to the tail sigma-algebra of the independent unit-time increments; it is a tail event. The Kolmogorov zero-one law gives $\mathbb{P}(A)\in \{0,1\}$. If its probability were one, then $B_n/\sqrt{n}\to 0$ in probability, contrary to its standard normal distribution at every $n$. Therefore $\mathbb{P}(A)=0$. Continuous-time convergence would imply integer-time convergence, soThe more general Brownian fluctuations exceed the square-root scale result is consistent with this direct tail-event proof. No independence of the normalized values themselves has been assumed.

The nonzero quadratic residues modulo five are $1,4$. The values of $x^3+x+1$ and the numbers of possible $y$ areThere are eight affine points and one point at infinity. Thus $\#E({\mathbb{F}}_5)=9$ and part(ii) gives $\boxed{\mathrm{tr}(\varphi )=5+1-9=-3}$. The curve is nonsingular, since its short-Weierstrass discriminant $-16(4+27)$ is nonzero modulo five.

The dual operation reverses products, so $\widehat{{\varphi }^2}=(\hat{\varphi }{)}^2$. Also $\varphi \hat{\varphi }=[5]$ and $\varphi +\hat{\varphi }=[-3]$. Squaring the latter identity givesTherefore $\boxed{\mathrm{tr}({\varphi }^2)=-1}$. Over ${\mathbb{F}}_{25}$ the Frobenius is ${\varphi }^2$, of degree $25$, soMore generally the Frobenius trace recurrence has $t_0=2$, $t_1=t$ and $t_j=t{t}_{j-1}-q{t}_{j-2}$; the square calculation is its first nontrivial case.

The $q$-power Frobenius isogeny sends affine coordinates to $(x^q,y^q)$, has degree $q$, and fixes exactly the points defined over ${\mathbb{F}}_q$, including the origin. ConsequentlyThe degree is $q$ because the $q$-power Frobenius on a smooth curve has function-field inseparable degree $q$. By the permitted separability of $1-\varphi$, this separable isogeny has exactly $\mathrm{deg}(1-\varphi )$ geometric kernel points; there is no inseparable multiplicity to remove. Part(i) therefore givesThis relates the elliptic-curve point count over a finite field to the integer trace, rather than treating the point count as an independently assumed definition of the trace.

Apply the dual isogeny degree identity to $1-\varphi$. Its dual is $1-\hat{\varphi }$, soRearrangement givesThe right side is multiplication by an integer, proving the assertion. In particular the trace of an elliptic-curve endomorphism isThe argument includes $\varphi =0$ and $\varphi =1$ with the zero-map convention. Multiplying $\hat{\varphi }=[t]-\varphi$ by $\varphi$ also yields the useful characteristic relation

Suppose an integer right triangle with square area exists, and choose one with the least positive hypotenuse $c$. It is primitive. Indeed, dividing all sides by their common divisor $d$ gives an integer right triangle whose area is the square of the rational number $s/d$, where $s^2$ was the original area. Its area is an integer, since one of its legs is even. A rational number whose square is an integer is itself an integer, as follows by writing it in lowest terms. Thus division by $d>1$ would give another square-area triangle with smaller hypotenuse.

By the primitive Pythagorean parametrization, after ordering the legs the triangle has sides $n^2-m^2,2nm,n^2+m^2$, with coprime $n>m>0$ of opposite parity. Its area isThese four positive factors are pairwise coprime: the pair $n-m,n+m$ has gcd dividing $2m$ and is odd, and the other pairs have gcd one by $\gcd (n,m)=1$. Since the product is a square, each factor is a square. WriteBoth $w,z$ are odd. Odd squares are $1(\mathrm{mod}8)$, so $2m=w^2-z^2$ is divisible by eight; hence $m$ is divisible by four and $v$ is even. Define positive integersThey satisfyThus $(r,s,u)$ is another positive integer right triangle with square area, but its hypotenuse $u=\sqrt{n}$ is strictly smaller than $n^2+m^2=c$. This contradicts minimality. No integer right triangle has a nonzero square area. This is the Fermat right triangle theorem, with the actual infinite descent construction exhibited.

Use both Kähler identities from the preceding parts, so that ${\partial }^{\ast }=i[\Lambda ,\bar{\partial }]$ and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. For odd operators $a,b$, write $\{a,b\}=ab+ba$. Direct expansion givesSince $\partial \bar{\partial }+\bar{\partial }\partial =0$, we obtainTogether with part (b), this proves the full Kähler Laplacian identity

Now take the exact form $\eta$. Being $\bar{\partial }$-exact, it is orthogonal to the harmonic space, so $H\eta =0$, and it is also $\bar{\partial }$-closed. Choose $\alpha =G\eta$ for the Dolbeault Green operator of the Dolbeault Laplacian. Commutation gives $\bar{\partial }\alpha =G\bar{\partial }\eta =0$. ThereforeSuppose also $\partial \eta =0$, and put $v={\bar{\partial }}^{\ast }\alpha$, $u=\partial v$. The mixed identities and $({\bar{\partial }}^{\ast }{)}^2=0$ giveThus $u=\partial {\bar{\partial }}^{\ast }\alpha$ is harmonic for the Dolbeault Laplacian. By the Kähler Laplacian identity it is also $\partial$-harmonic and hence ${\partial }^{\ast }u=0$. But it is $\partial$-exact, soConsequently $\boxed{\partial {\bar{\partial }}^{\ast }\alpha =0}$: the form $v$ is $\partial$-closed.

For every harmonic form $h$ of the same bidegree as $v$, $⟨v,h⟩=⟨\alpha ,\bar{\partial }h⟩=0$. Harmonic spaces for the two Dolbeault Laplacians coincide, so $v$ has no $\partial$-harmonic component. Apply the $\partial$ version of the Dolbeault Hodge decomposition. Since $v$ is $\partial$-closed, it has no ${\partial }^{\ast }$-coexact component either. Thus $v=\partial \varphi$, with $\varphi \in A^{p-1,q-1}$. We conclude the requested ddbar lemma:For example $\varphi ={\partial }^{\ast }Gv$ is a choice, since the same Green operator serves ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. If $p=0$ or $q=0$, the relevant negative-bidegree space is zero and the argument forces $\eta =0$. Compactness, the absence of boundary and the Kähler condition are essential to this global harmonic argument.

Complex conjugation preserves the real Kähler form and interchanges $\partial ,\bar{\partial }$ and their formal adjoints. Conjugating the permitted identity among the Kähler identities therefore yields $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$. The same nilpotence calculation as in part (a) gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$. Now expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$:Both mixed anticommutators vanish, proving $\boxed{{\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}}$.

With the usual exterior-boundary interpretation that the harmonic function remains bounded as $r\to \infty$, positive radial powers and $\log r$ are excluded. HenceThe series must converge locally for $r>1$, with any required finite boundary values at $r=1$. These are the interior harmonic Fourier expansions in planar concentric domains after the inversion $r\mapsto 1/r$.

If “finite” means only finite at each finite point, without a condition at infinity, growing solutions such as $r\cos \theta$ and $\log r$ are allowed. Under that literal reading the general exterior expansion iswith local convergence. Boundedness at infinity is the condition selecting the decaying exterior expansion.

Set $F(z)=\exp (ig(z{)}^2)$. Since $\mathrm{Im}(g^2)=2uv$, its modulus is $e^{-2uv}\le 1$. The Liouville theorem again makes $F$ constant, and $F^{\prime }=2ig{g}^{\prime }F=0$ implies $(g^2{)}^{\prime }=0$. Hence $g^2$ is a constant $c$. If $c=0$, then $g=0$. If $c\ne 0$, then $g$ never vanishes and $g{g}^{\prime }=0$ implies $g^{\prime }=0$. In either case $g$ is constant. This is also an application of the one-sided product bound for an entire function after rotating $g$ by $i$.