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In an abelian category, for $f:A\to B$ set $\mathrm{coim}f=\mathrm{coker}(\ker f)$ and $\mathrm{im}f=\ker (\mathrm{coker}f)$. The canonical map from coimage to image is an isomorphism. Thus the image factorization in an abelian category isunique up to a unique compatible isomorphism.

The short five lemma states that in a commuting diagram of short exact sequences $0\to L\to M\to R\to 0$ and $0\to L^{\prime }\to M^{\prime }\to R^{\prime }\to 0$ in an abelian category, if the vertical maps $L\to L^{\prime }$ and $R\to R^{\prime }$ are isomorphisms, then so is $M\to M^{\prime }$.

For the Five lemma, consider commuting exact rowswith vertical maps $v_i:A_i\to B_i$. We prove the usual stronger version: $v_1$ epic, $v_2$ and $v_4$ invertible, and $v_5$ monic imply that $v_3$ is invertible. In particular the conclusion holds if all four outer vertical maps are isomorphisms.

Put $L=\mathrm{im}{d}_2$ and $R=\mathrm{im}{d}_3$, and define $L^{\prime },R^{\prime }$ similarly. Exactness gives short exact sequencesThe functoriality of abelian image factorization supplies their vertical outer maps. We show that both are isomorphisms.

First, $L\cong \mathrm{coker}{d}_1$ and $L^{\prime }\cong \mathrm{coker}{d}_1^{\prime }$: exactness at $A_2$ identifies $\mathrm{im}{d}_1$ with $\ker d_2$, and the image-coimage isomorphism for $d_2$ gives the claimed cokernel. Write $q:A_2\to L$ and $q^{\prime }:B_2\to L^{\prime }$ for these categorical cokernels. To construct an inverse to the induced map $\ell :L\to L^{\prime }$, observeCancellation of the epimorphism $v_1$ gives $q{v}_2^{-1}{d}_1^{\prime }=0$, so $q{v}_2^{-1}$ factors uniquely as $\bar{\ell }{q}^{\prime }$ with $\bar{\ell }:L^{\prime }\to L$. From $\ell q=q^{\prime }{v}_2$, composing with the epic $q,q^{\prime }$ gives $\bar{\ell }\ell =1_L$ and $\ell \bar{\ell }=1_{L^{\prime }}$.

Second, $R\cong \ker d_4$ and $R^{\prime }\cong \ker d_4^{\prime }$ by exactness at the fourth objects. Write their inclusions as $j:R\to A_4$ and $j^{\prime }:R^{\prime }\to B_4$. The equality $d_4^{\prime }{v}_4=v_5{d}_4$ gives the induced map $\rho :R\to R^{\prime }$. Sinceand $v_5$ is a monomorphism, $d_4{v}_4^{-1}{j}^{\prime }=0$. Kernel universality gives $\bar{\rho }:R^{\prime }\to R$ with $j\bar{\rho }=v_4^{-1}{j}^{\prime }$. Cancelling the monic $j,j^{\prime }$ proves $\bar{\rho }\rho =1_R$ and $\rho \bar{\rho }=1_{R^{\prime }}$.

Apply the short five lemma to the two short exact sequences: their outer maps $\ell ,\rho$ are invertible, so $v_3$ is an isomorphism. This five lemma via image factorization argument uses only universal properties and therefore works in any abelian category, without treating its objects as literal sets of elements.

Put $C={\bigcup }_{r\ge 1}\mathbb{Q}({\mu }_{p^r})$, where ${\mu }_{p^r}$ is the group of $p^r$th roots of unity. The compatible Galois groups of the cyclotomic fields giveFor odd $p$, the Teichmüller character givesThe p-adic logarithm identifies $1+p{\mathbb{Z}}_p$ with $p{\mathbb{Z}}_p$, which is topologically isomorphic to ${\mathbb{Z}}_p$. More explicitly, $a\mapsto (1+p{)}^a$ is an isomorphism from the additive p-adic integers onto $1+p{\mathbb{Z}}_p$: the logarithm of $1+p$ has valuation one and generates $p{\mathbb{Z}}_p$. Take the fixed field of ${\mu }_{p-1}$ in $C$.

For $p=2$, use insteadThe p-adic logarithm identifies the second factor with $4{\mathbb{Z}}_2$, and $a\mapsto 5^a$ identifies it with ${\mathbb{Z}}_2$. The fixed field of $\{\pm 1\}$ in $C$ is the union of the maximal real subfields of the cyclotomic fields of $2$-power conductor. Thus in both cases the cyclotomic Zp-extension exists:where ${\Delta }_p={\mu }_{p-1}$ for odd $p$ and ${\Delta }_2=\{\pm 1\}$.

The closed subgroup $p^n{\mathbb{Z}}_p$ is the unique subgroup of index $p^n$, so the Galois correspondence supplies the unique degree-$p^n$ field $\mathbb{Q}[p^n]$. In the odd case it is the ${\Delta }_p$-fixed field in $\mathbb{Q}({\mu }_{p^{n+1}})$; in the even case it is the real subfield of $\mathbb{Q}({\mu }_{2^{n+2}})$. These finite layers are all totally real number fields.

Now set $A=\mathbb{Q}[p^n]$, $B=\mathbb{Q}[p^m]$, and let $H_A$ be the ordinary Hilbert class field of $A$. Thus $H_A/A$ is abelian, unramified at finite primes, and split at every real place, with degree $h(A)$. If $m=n$ there is nothing to prove. Otherwise the unique prime over $p$ is totally ramified in $B/A$, by the permitted total-ramification fact and multiplicativity of the ramification index. Any nontrivial intermediate field of $B/A$ is ramified at that prime. HenceIt follows that the two extensions are linearly disjoint field extensions, andUnramified extensions remain unramified after base change. The real places also remain split: $H_A$ and $B$ are totally real, so their compositum is totally real. Therefore $B{H}_A/B$ lies inside the ordinary Hilbert class field $H_B/B$. Its degree divides $[H_B:B]=h(B)$, provingThis is class number divisibility with split real places; using a class-field statement restricted to imaginary fields would unnecessarily omit the present real layers.

Let $q:{\mathbb{CP}}^8\to Q$ collapse the indicated Complex projective space. Write $x\in H^2({\mathbb{CP}}^8;\mathbb{Z})$ for the positive generator, so the cohomology ring of complex projective space is $\mathbb{Z}[x]/(x^9)$. The inclusion of ${\mathbb{CP}}^3$ sends $x$ to its corresponding generator and is an isomorphism in degrees $0,2,4,6$.

The CW complex structure makes the inclusion a cofibration, so the positive-degree cohomology of $Q$ identifies with the relative cohomology of the pair. Its long exact sequence shows that $q^{\ast }$ is injective in positive degrees, with image the additive subgroup generated by $x^4,x^5,x^6,x^7,x^8$. Equivalently, the surviving cellular cochain complex has one copy of $\mathbb{Z}$ in each of degrees $8,10,12,14,16$ and zero differential. Let $u_i$ denote the class with $q^{\ast }{u}_i=x^i$, for $4\le i\le 8$.

Naturality of the cup product, together with the injectivity of $q^{\ast }$, now determines all multiplication:Every other product of positive-degree basis elements is zero. Indeed, their degrees exceed $16$. In particular, the surviving classes do not form a polynomial algebra on a degree-eight generator: the independent classes in degrees $10,12,14$ must also be retained. This is an instance of the cohomology ring of a collapsed projective subspace.

Write $A=S^2\times S^2$ and let $i:A\to X$ be its inclusion after the attachment. The cohomology ring of a product of two spheres iswhere $a,b$ come from the first and second factors and $ab$ is the chosen orientation class. The diagonal pulls both $a$ and $b$ back to the same generator of $H^2(S^2;\mathbb{Z})$.

There is one new three-cell. In the cellular chain complex, its boundary has coordinates $(1,1)$ in the two-dimensional cells. Thus the relevant differential isand the original four-cell still has zero boundary. This gives $H_2(X;\mathbb{Z})=\mathbb{Z}$, $H_3(X;\mathbb{Z})=0$, $H_4(X;\mathbb{Z})=\mathbb{Z}$, and no other positive homology groups. Equivalently, the relative cohomology sequence of $(X,A)$ gives an injective restriction in degree two with image $\mathbb{Z}(a-b)$, and an isomorphism in degree four.

Choose $u\in H^2(X;\mathbb{Z})$ and $v\in H^4(X;\mathbb{Z})$ by $i^{\ast }u=a-b$, $i^{\ast }v=ab$. Naturality of the cup product givesSince restriction is injective in degree four, $u^2=-2v$. All further positive-degree cup products vanish by dimension. ThereforeChanging the sign of $v$ would instead give $u^2=2v$; the displayed sign uses the product orientation $ab$. The factor $2$ is the characteristic feature of a cup square after a diagonal sphere attachment.

First check the signs directly. For $(x,y)\in M_i$, applying the proposed differential twice givesbecause $f$ is a chain map. Thus $M$ is a chain complex.

Let $C^{\prime }$ be the shifted chain complex with $C_i^{\prime }=C_{i-1}$ and differential $d_C$; here there is no minus sign in that differential. Inclusion in the second summand and projection onto the first give a short exact sequence of chain complexesTo compute the connecting map in its long exact sequence in homology, lift a cycle $x\in C_{i-1}$ to $(x,0)\in M_i$. Its boundary is $(0,(-1{)}^{i}f(x))$. Hence the connecting map $H_i(C^{\prime })\to H_{i-1}(D)$ is $(-1{)}^i{f}_{\ast }$, and the relevant exact portion isIf every $f_{\ast }$ is an isomorphism, exactness makes $H_i(M)=0$. Conversely, if every $H_i(M)$ is zero, the adjacent exact portions make every $f_{\ast }$ both injective and surjective. ThereforeThis is the acyclicity criterion for a mapping cone, with the degree-dependent signs adjusted to the given convention.

For the assertion about spaces, use the cellular approximation theorem to replace $f$ by a cellular map, and take the induced map of the finite free cellular chain complexes. Tensoring these complexes and their cone with ${\mathbb{F}}_p$ gives the corresponding mod-$p$ complexes and cone. The same exact-sequence argument works over ${\mathbb{F}}_p$, so the assumed homology isomorphisms implyThe universal coefficient theorem for homology givesIn particular $H_i(M)\otimes {\mathbb{F}}_p=0$ for every prime. Each $H_i(M)$ is a finitely generated abelian group. A nonzero free summand would survive tensoring with every ${\mathbb{F}}_p$, while a nonzero finite cyclic summand would survive for a prime dividing its order. Thus $H_i(M)=0$ in every degree, by detection of integral acyclicity modulo primes. Applying the cone criterion once more proves the integral homology map is an isomorphism in every degree. Finite generation is what makes detection by all prime fields sufficient.

Work first with coefficients ${\mathbb{F}}_2$. For a rank-$r$ real vector bundle $E\to B$ over a CW complex, choose a fibre metric and its disk bundle $D(E)$ and sphere bundle $S(E)$. The mod-two Thom class is the unique class $U\in H^r(D(E),S(E);{\mathbb{F}}_2)$ restricting to the nonzero generator on every fibre pair. The Thom isomorphism theorem asserts thatNo orientation of the vector bundle is needed with these coefficients.

Let $j:H^{\ast }(D(E),S(E))\to H^{\ast }(D(E))$ forget the relative condition, and let $s:B\to D(E)$ be the zero section. Set $e_2(E)=s^{\ast }j(U)$, the top Stiefel–Whitney class $w_r(E)$. Since $D(E)$ retracts onto the zero section, $j(U)={\pi }^{\ast }{e}_2(E)$. Thus the relative-to-absolute map corresponds under the Thom isomorphism theorem to multiplication by $e_2(E)$. Substituting these identifications into the long exact sequence in relative cohomology of $(D(E),S(E))$ gives the unoriented Gysin sequencewhere $p$ is the sphere bundle projection.

Apply this to the real tautological line bundle ${\gamma }_n\to {\mathbb{RP}}^n$. Its sphere bundle is $S^n$, with projection the antipodal double cover, and put $t=e_2({\gamma }_n)\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$. Suppose $n\ge 1$. The map on $H^0$ from the connected base to the connected sphere is an isomorphism, so exactness shows that multiplication by $t$ is injective from $H^0$ to $H^1$. Since both groups are one-dimensional, $t$ is a generator.

For $1<j<n$, the adjacent sphere cohomology groups in the Gysin sequence vanish, so multiplication by $t$ is an isomorphism from degree $j-1$ to degree $j$. If $n>1$, the top portion isThe last nonzero arrow is surjective between one-dimensional groups, hence an isomorphism; the preceding arrow is zero, so multiplication by $t$ is again an isomorphism. For $n=1$ the earlier $H^0$ argument already gives the top multiplication. Thus $1,t,\dots ,t^n$ are the nonzero generators in their respective degrees, while $t^{n+1}=0$ by dimension. We obtainFor $n=0$ this says simply $H^{\ast }({\mathbb{RP}}^0;{\mathbb{F}}_2)={\mathbb{F}}_2$, with $t=0$. This computes the mod-two cohomology ring of real projective space, including its multiplication rather than only its additive groups.

Now use integral coefficients and an oriented rank-$n$ vector bundle. Its orientation selects an integral Thom class $U$. Define its Euler class byAgain $j(U)={\pi }^{\ast }e(E)$. The cup product of two relative classes in $H^{\ast }(D(E),S(E))$ agrees with the mixed relative/absolute product after forgetting the relative condition on either factor. ConsequentlyThis is the cup square of a Thom class. If $n$ is odd, graded commutativity of the cup product gives $U\cup U=-U\cup U$, hence $2U^2=0$. Since $U\cup {\pi }^{\ast }(2e(E))$ is, up to the graded sign, the Thom isomorphism image of $2e(E)$, its vanishing impliesThus the Euler class of an oriented odd-rank vector bundle is two-torsion; the conclusion is integral and does not require the base cohomology to be torsion-free.

Use the covariant Yoneda embedding $Y(A)=\mathcal{C}(A,-)$. An arrow $f:B\to A$, viewed as an arrow $A\to B$ in the opposite category, induces precomposition $h\mapsto hf$. Identities and composition are preserved because composition in $\mathcal{C}$ is associative.

The covariant Yoneda lemma is the bijectionIts inverse sends $x\in F(A)$ to ${\alpha }_B(h)=F(h)(x)$. This is a natural transformation: for $k:B\to C$, functoriality gives $F(k){\alpha }_B(h)=F(kh)(x)={\alpha }_C(kh)$. Conversely, naturality of an arbitrary $\alpha$ at $h:A\to B$ gives ${\alpha }_B(h)=F(h){\alpha }_A(1_A)$. Evaluation at the identity morphism therefore makes the two constructions inverse. Their formulas also prove naturality in $F$ and, contravariantly, in $A$.

Taking $F=Y(B)$ identifies natural transformations $Y(A)\to Y(B)$ with $\mathcal{C}(B,A)$. Thus $Y$ is full and faithful, and reflects isomorphisms; two image objects are isomorphic exactly when the original objects are isomorphic. It also carries existing colimits in $\mathcal{C}$ to categorical limits of covariant representable functors: maps out of a colimit are exactly compatible families of maps out of its diagram. Equivalently, this embedding preserves existing categorical limits in ${\mathcal{C}}^{\mathrm{op}}$. The direction matters: the paper uses covariant representables, rather than the more usual contravariant Yoneda embedding.

A representation of a functor $F$ is $(A,x)$ with $x\in F(A)$ such that $h\mapsto F(h)(x)$ is a bijection $\mathcal{C}(A,B)\to F(B)$ for every $B$. These bijections are natural by functoriality. The element $x$ is its universal element.

If $(B,y)$ is another representation, universality gives unique arrows $f:A\to B$ and $g:B\to A$ satisfying $F(f)x=y$ and $F(g)y=x$. Consequently $F(gf)x=x=F(1_A)x$, so $gf=1_A$ by injectivity of the representing bijection. Similarly $fg=1_B$. Therefore $f$ is the unique compatible isomorphism. This proves uniqueness of functor representations; uniqueness refers to an isomorphism carrying the specified universal elements to one another, rather than to every isomorphism between the underlying objects.

An initial object by itself is a weakly initial set. For the converse, let $(W_i{)}_{i\in I}$ be a small weakly initial family and form its product in a category $W$. Given $X$, choose an arrow $W_i\to X$; its composite with the projection $W\to W_i$ shows that $W$ is weakly initial.

Local smallness makes $\mathrm{End}(W)$ a set, so completeness supplies the simultaneous equalizer $e:E\to W$ of all endomorphisms of $W$ with $1_W$. Thus $he=e$ for every $h\in \mathrm{End}(W)$. The object $E$ is weakly initial because it maps to $W$.

For parallel arrows $a,b:E\to X$, take their equalizer $j:Y\to E$. Weak initiality of $W$ supplies $t:W\to Y$. The endomorphism $ejt$ of $W$ satisfies $(ejt)e=e$, and monicity of $e$ gives $jte=1_E$. Therefore $j$ is both a monomorphism and a split epimorphism, hence an isomorphism. Since $aj=bj$, we obtain $a=b$. There is already at least one arrow from $E$ to every $X$, so $E$ is initial. This proves the initial-object lemma for complete categories with a weakly initial set, with smallness used exactly where the endomorphisms are equalized.

For the monad $(T,\eta ,\mu )$, an algebra for a monad is $(A,a:TA\to A)$ with $a{\eta }_A=1_A$ and $aT(a)=a{\mu }_A$. A morphism of algebras for a monad $f:(A,a)\to (B,b)$ satisfies $fa=bTf$. These objects and arrows form the Eilenberg-Moore category ${\mathcal{C}}^T$.

The free algebra functor sends $A$ to $(TA,{\mu }_A)$ and $f$ to $Tf$. The monad identities verify the algebra laws. For the forgetful functor $U$, the free adjunction iswith inverse $v\mapsto bTv$. The algebra law makes the latter an monad algebra morphism. The identities $h{\eta }_A\mapsto bT(h{\eta }_A)=h{\mu }_{A}T{\eta }_A=h$ and $bTv{\eta }_A=b{\eta }_{B}v=v$ prove the bijection.

In the category of adjunctions inducing a fixed monad, objects are adjunctions $F⊣G$ with their induced monad identified with $T$. A morphism to $F^{\prime }⊣G^{\prime }$ is a functor $H$ between the right-hand categories satisfying $G^{\prime }H=G$, $HF=F^{\prime }$ and compatibility with units and counits. With these strict identifications, defineThe triangle identities give the unit algebra law; naturality of $\epsilon$ at ${\epsilon }_B$ gives the multiplication law. Naturality at $f$ makes $Gf$ an monad algebra morphism. Moreover $UK=G$, $KF=F^T$ because $G{\epsilon }_{FA}={\mu }_A$, and $K{\epsilon }_B$ is the free-adjunction counit at $K(B)$. Hence $K$ is a morphism into the Eilenberg-Moore adjunction.

For any other such $H$, its underlying object at $B$ must be $GB$. Counit compatibility forces its algebra action to be $G{\epsilon }_B$, and the forgetful functor, which is a faithful functor, forces $H(f)=Gf$. Thus $H=K$. This proves terminality of the Eilenberg-Moore adjunction. If adjunctions are specified only up to coherent isomorphisms, the same argument gives uniqueness up to the corresponding compatible natural isomorphism.

The crude monadicity theorem in its reflexive-coequalizer form says: if $F⊣G:\mathcal{D}\to \mathcal{C}$, $G$ is a conservative functor, $\mathcal{D}$ has reflexive coequalizers, and $G$ preserves them, then the Eilenberg-Moore comparison functor $K:\mathcal{D}\to {\mathcal{C}}^T$, for $T=GF$, is an equivalence of categories. Requiring all coequalizers to exist and be preserved is a stronger sufficient form.

For an algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is a reflexive pair with common section $F{\eta }_A$: both composites are the identity by the algebra unit law and the triangle identity. An arrow $u:FA\to B$ transposes to $v=Gu{\eta }_A:A\to GB$. The two composites $uFa$ and $u{\epsilon }_{FA}$ transpose to $va$ and $G{\epsilon }_{B}Tv$, respectively. Indeed the latter transpose is $Gu{\mu }_A{\eta }_{TA}=Gu$, while $G{\epsilon }_{B}Tv=Gu{\mu }_{A}T{\eta }_A=Gu$ by counit naturality. Thus $u$ equalizes the pair exactly when $v$ is an monad algebra morphism $(A,a)\to K(B)$. The coequalizer property gives natural bijectionsThese define $L$ on arrows by uniqueness, so $L⊣K$. Its unit has underlying arrow $\beta =Gq{\eta }_A:A\to GL(A,a)$.

By preservation, $Gq$ coequalizes $Ta$ and ${\mu }_A$ in $\mathcal{C}$. The action $a:TA\to A$ is a split coequalizer of that pair: take $s={\eta }_A$ and $t={\eta }_{TA}$, with ${\mu }_{A}t=1_{TA}$ and $Tat={\eta }_{A}a$. Thus there is a unique isomorphism $\alpha :GL(A,a)\to A$ with $\alpha Gq=a$. We have $\alpha \beta =a{\eta }_A=1_A$. AlsoConsequently $\beta \alpha Gq=Gq$, and epimorphic cancellation gives $\beta \alpha =1$. The adjunction unit $\beta$ is an monad algebra morphism; its invertible underlying arrow has an algebra-morphism inverse. Thus the unit of $L⊣K$ is an isomorphism.

For its counit $c_B:LK(B)\to B$, the triangle identity gives $K(c_B){\beta }_{K(B)}=1_{K(B)}$. Hence $K(c_B)$, and therefore $G(c_B)$, is an isomorphism. Since $G$ is a conservative functor, $c_B$ is an isomorphism too. Both unit and counit of $L⊣K$ are invertible, which proves the claimed equivalence. This proof uses only reflexive coequalizers in $\mathcal{D}$ and split coequalizers in $\mathcal{C}$.