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A map that is both a model fibration and a quasi-isomorphism. The homology isomorphism together with degree-one surjectivity forces surjectivity in degree zero. Equivalently it has the right lifting property against the generating cofibrations for nonnegative chain complexes.

The category $D^{+}(X)$ is obtained from bounded below cochain complexes of sheaves by passing to cochain-homotopy classes of maps and then inverting quasi-isomorphisms. It can equally be modeled by injective resolutions. The cohomology-sheaf functors descend to this localization, and an object is zero exactly when all its cohomology sheaves vanish.

A commutative differential graded algebra has multiplication $ab=(-1{)}^{|a||b|}ba$ and a degree-one differential with $d^2=0$ and $d(ab)=d(a)b+(-1{)}^{|a|}ad(b)$. Over $\mathbb{Q}$, its free algebra on graded generators is polynomial on even-degree generators and exterior on odd-degree generators. Rational rational polynomial differential forms on a space supply such an algebra, modelled up to quasi-isomorphism by a Sullivan minimal model.

An exact, or distinguished, triangle is one isomorphic to $E\stackrel{f}{\to }F\to \mathrm{Cone}(f)\to E[1]$. For a degreewise short exact sequence $0\to E\to F\to G\to 0$, the map $\mathrm{Cone}(f)\to G$ taking $(v,e)$ to the image of $v$ is a quasi-isomorphism: its kernel is the contractible cone of the identity on $E$. Consequently the short exact sequence defines such a triangle.

A space is formal over $\mathbb{Q}$ if its rational cochain commutative differential graded algebra is joined to $(H^{\ast }(X;\mathbb{Q}),0)$ by quasi-isomorphisms. Equivalently in the usual simply connected Sullivan-model setting, its minimal model has a quasi-isomorphism to its cohomology algebra with zero differential. This says its rational homotopy type can be recovered from the cohomology algebra; zero cup products alone do not imply formality of a topological space without an additional argument.

A morphism in the chosen weak-equivalence class of a model category. In the projective model structure on nonnegative chain complexes these are quasi-isomorphisms.

For a cochain complex $E$, define the shift of a cochain complex byFor a cochain map $a:E\to F$, put $(\Sigma a{)}^i=a^{i+1}$. This is again a cochain map, so the construction is a functor. The sign ensures the standard cone identities; the shifted differential squares to zero. Iteration gives $E[r{]}^i=E^{i+r}$ and differential $(-1{)}^r{d}_E$, and $H^i(E[r])=H^{i+r}(E)$.

For $\varphi :E\to F$, its algebraic mapping cone isApplying this differential twice gives first component $d_F^{2}f+d_F\varphi e-\varphi d_{E}e=0$, since $\varphi$ is a cochain map, and second component $d_E^{2}e=0$. The mapsare cochain maps: $d_{C}u(f)=(d_{F}f,0)=u(d_{F}f)$ and $v(d_C(f,e))=-d_{E}e=d_{\Sigma E}v(f,e)$. They are maps of complexes of sheaves, because every displayed operation is a sheaf morphism in each degree.

The bounded below derived category of sheaves $D(X)=D^{+}(X)$ can be constructed by first quotienting the category of bounded below cochain complexes by cochain homotopy and then formally inverting all quasi-isomorphisms. A morphism may be represented by a roof $E\stackrel{\sim }{\leftarrow }{E}^{\prime }\to F$. Alternatively, take injective resolutions of sheaves and use homotopy classes of maps between the resulting injective complexes. The cohomology sheaf functors descend to the localization, because they already identify homotopic maps and turn quasi-isomorphisms into isomorphisms. If $E\cong 0$ in $D(X)$, applying these functors gives ${\mathcal{H}}^i(E)=0$ for every $i$. Conversely, if all these sheaves vanish, the unique map $E\to 0$ is a quasi-isomorphism, which becomes invertible in $D(X)$. HenceAn exact triangle in a derived category is a triangle isomorphic to a cone triangleIn particular the last arrow is part of the structure; a mere sequence of three objects with successive composite zero does not suffice.

Suppose $0\to E\stackrel{\varphi }{\to }F\stackrel{\psi }{\to }G\to 0$ is degreewise exact. DefineThen $r(d_C(f,e))=\psi (d_{F}f)+\psi \varphi (e)=d_G\psi (f)=d_{G}r(f,e)$, so $r$ is a cochain map. It is natural under maps of short exact sequences. It is degreewise surjective as a sheaf map, and its kernel can be identified with $E^i\oplus E^{i+1}$ using $(a,e)\mapsto (\varphi (a),e)$. In these coordinates the kernel differential isThere is a contracting homotopy $h^i(a,e)=(0,a)$ in degree $i-1$:Thus the kernel has zero cohomology sheaves. The cohomology-sheaf exact sequence is the sequence of the short exact sequence of complexes $0\to \ker r\to C(\varphi )\to G\to 0$, or equivalently its ordinary complex-cohomology sequence at every stalk. It follows that $r$ induces isomorphisms of all cohomology sheaves, provingUnder this isomorphism the cone triangle becomes the exact trianglewhere $\delta =v\circ r^{-1}$ in the derived category.

For the final construction write $U=S^2\setminus S^1$ and set $A=i_{!}{\mathbb{Q}}_U$, $B={\mathbb{Q}}_{S^2}$, $C=j_{\ast }{\mathbb{Q}}_{S^1}$, each initially placed in degree zero. The stated sequence $0\to A\to B\to C\to 0$ is exact on stalks: on $U$ its first map is an isomorphism and its last sheaf is zero; on the equator its first sheaf is zero and the second map is an isomorphism. Take its connecting morphismThe source has only ${\mathcal{H}}^0(C)=C$, whereas the target has only ${\mathcal{H}}^{-1}(A[1])=A$. Therefore every induced map ${\mathcal{H}}^i(\mu )$ is zero: in any degree at least one of its source and target is zero. This is a ghost morphism in a derived category.

To prove the derived morphism itself is nonzero, apply derived global sections, giving the hypercohomology sequence. The extension by zero identifies $H^{\ast }(S^2;A)$ with the compactly supported cohomology of $U$, and closed direct image identifies $H^{\ast }(S^2;C)$ with $H^{\ast }(S^1;\mathbb{Q})$. The complement $U$ is two open disks, each homeomorphic to ${\mathbb{R}}^2$, soThe degree-one part of the resulting exact sequence isThus ${\mu }_{\ast }$ is injective, in particular nonzero. With the hemispheres oriented from $S^2$, the last map is $(a,b)\mapsto a+b$, so an appropriately oriented equator generator maps to $(1,-1)$. A zero morphism in the derived category of sheaves would induce a zero map after derived global sections and taking cohomology. Consequently $\mu \ne 0$, although ${\mathcal{H}}^i(\mu )=0$ for every $i$. This establishes explicitly that the collection of cohomology-sheaf functors detects zero objects but is not faithful on morphisms.

It is enough to assume that $h_k:{\pi }_k(X)\otimes \mathbb{Q}\to H_k(X;\mathbb{Q})$ is surjective for every $k\ge 2$; an integrally surjective Hurewicz homomorphism certainly has this property. First give the minimal-model argument suggested by the hint. Let $A=(\Lambda V,d)$ be a simply connected Sullivan minimal model, with $V$ in degrees at least two. Put $A^{+}={⨁}_{k>0}{A}^k$. Its indecomposable quotient of an augmented algebra isMinimality makes the induced differential on $V$ zero, so projection is a chain map $q:A^{+}\to (V,0)$. The Sullivan model identification says that the induced mapis dual to rational Hurewicz homomorphism. Under the usual finite-type duality convention, the assumed surjectivity makes $q_{\ast }$ injective.

A product of two positive cohomology classes has a decomposable cocycle representative, so its image under $q_{\ast }$ is zero. Injectivity therefore givesThis deduction uses the hypothesis; zero cup products by themselves would not yet prove formality of a topological space.

Choose a homogeneous cohomology basis $h_{\lambda }$ and cocycle representatives $z_{\lambda }\in A$. Their indecomposable parts $q(z_{\lambda })$ are linearly independent. Extend them to a graded basis of $V$ and replace the corresponding generators by the actual cocycles $z_{\lambda }$. This is a valid change of free algebra generators: each $z_{\lambda }$ differs from its indecomposable part by products of generators of strictly smaller degree, so degree induction constructs the inverse substitution. Call these closed generators $C$, and the remaining generators $N$.

Define a map of graded algebrasIts target has zero differential. Every differential in $A$ is decomposable, and all products of positive target classes vanish, so $\varphi$ is a chain map. If $z$ is any positive cocycle, its indecomposable part is in the span of the $q(z_{\lambda })$, because these form the image of $q_{\ast }$. Write $q(z)={\sum }_{\lambda }{c}_{\lambda }q(z_{\lambda })$. Then $\varphi (z)={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$, while injectivity of $q_{\ast }$ gives $[z]={\sum }_{\lambda }{c}_{\lambda }{h}_{\lambda }$. Thus $\varphi$ induces the identity on cohomology and is a quasi-isomorphism. This proves that $X$ is a formal space.

The rational wedge sum assertion has a direct topological proof, which also avoids finite-dimensional duality restrictions in the preceding model argument. Choose a homogeneous basis $b_{\lambda }$ of ${\tilde{H}}_{\ast }(X;\mathbb{Q})$. By surjectivity, each basis class is the image under the Hurewicz homomorphism of an element of ${\pi }_{n_{\lambda }}(X)\otimes \mathbb{Q}$. In $X_{\mathbb{Q}}$ this element is represented by an actual sphere map $S^{n_{\lambda }}\to X_{\mathbb{Q}}$. Simple connectivity and $H_1=0$ ensure $n_{\lambda }\ge 2$. The wedge sum of these maps givesand induces an isomorphism of rational homology by the choice of basis. The rational Whitehead theorem now makes it a rational homotopy equivalence, givingA wedge sum of spheres is a formal space. For each individual sphere, its augmented minimal model maps by a quasi-isomorphism to its sphere cochain algebra, and also to its cohomology algebra, by killing the extra odd generator in the even-sphere case. Take based simplicial models of the spheres. Their wedge sum is a simplicial pushout over the common basepoint, so the algebra of rational polynomial differential forms is the corresponding fibre product of augmented sphere algebras. Taking fibre products of these model maps gives a quasi-isomorphism zigzag to the square-zero cohomology algebra. This also works for infinitely many sphere summands: the positive parts are products of the individual complexes, and products of vector-space exact sequences over a field are exact. In the finite case its minimal resolution is the minimal model of a wedge of simply connected spheres. Thus the topological proof establishes both conclusions without imposing finite rank on the sphere basis. For the zero reduced-homology case the wedge sum is a point. Both the minimal-model construction and the explicit sphere map explain why the surjectivity hypothesis is decisive.

Let $u_n$ be the rationalized universal class of degree $n$. The rational cohomology of an integral Eilenberg–MacLane space isHere $\Lambda$ is an exterior algebra. For $n=1$, $K(\mathbb{Z},1)\simeq S^1$. The standard path-loop spectral-sequence calculation supplies the induction: inthe total space is contractible and the fundamental fiber class transgresses to $u_n$. An odd exterior fiber generator gives an even polynomial base generator. An even polynomial fiber generator gives an odd exterior base generator; the differential on its $k$th power has coefficient $k$, which is invertible over $\mathbb{Q}$. The multiplicative spectral sequence then has no remaining positive-degree classes in the total space. This is the rational transgression calculation for Eilenberg–MacLane spaces; it includes the absence of additional base generators.

For $m=2n+1\ge 3$, choose $S^m\to K(\mathbb{Z},m)$ representing the integral fundamental class. The ring calculation shows that this map is a rational homology equivalence: both spaces have rational cohomology only in degrees zero and $m$. The rational Whitehead theorem for simply connected spaces identifies their rational homotopy groups. Since the target has only ${\pi }_m=\mathbb{Z}$, the rational homotopy groups of a sphere in odd dimension areFor $n=0$, this follows directly from ${\pi }_1(S^1)=\mathbb{Z}$ and the contractible universal cover of the circle, which makes every higher homotopy group zero.

Let $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$, with the standard complex orientations. It is simply connected by the Seifert-van Kampen theorem applied to the punctured summands. Classes $a,b\in H^2(M;\mathbb{Q})$ can be chosen from the two summands. Their cross product vanishes, while their squares equal the oriented top class:Thus the cohomology ring of the connected sum of two complex projective planes isThe two relations also kill all cubic monomials, so its dimensions are $1,2,1$ in degrees $0,2,4$ and zero otherwise.

We use the Sullivan minimal model dictionary: for a simply connected finite-type space, the dual of its degree-$i$ generator space is ${\pi }_i(M)\otimes \mathbb{Q}$. The following free graded-commutative differential algebra is the Sullivan model of the connected sum of two complex projective planes:It is minimal because all differentials of generators are decomposable.

To verify that no further generators are required, observe that $ab,a^2-b^2$ is a regular sequence in $\mathbb{Q}[a,b]$. The first polynomial is a nonzerodivisor. If $(a^2-b^2)h$ is divisible by $ab$, restricting to each coordinate axis forces $h$ to vanish on both axes, hence to be divisible by $ab$. The second polynomial is therefore a nonzerodivisor modulo the first. The Koszul complex of this regular sequence is exactly $\mathcal{M}$, so its cohomology is the quotient ring above, with no additional odd cohomology.

For completeness, choose rational polynomial forms representing $a,b$ on $M$. Their product and the difference of their squares are exact; choose degree-three primitives for them. Sending $r,s$ to those primitives defines a differential-algebra map from $\mathcal{M}$ to the rational polynomial forms on $M$. It induces the specified cohomology-ring isomorphism and hence is a quasi-isomorphism. This verifies the model directly, rather than assuming that a cohomology presentation alone automatically determines all rational homotopy.

The model has exactly two degree-two and two degree-three generators. Consequently the rational homotopy groups of the connected sum of two complex projective planes are

For a cohomological spectral sequence, $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$ and the next page is its cohomology. In the bounded setting, convergence of a spectral sequence to $H^{\ast }$ means that each $H^n$ has a finite decreasing, exhaustive and separated filtration withHere $E_{\infty }^{p,q}$ is the eventual stable value. This determines the associated graded module of $H^n$, rather than automatically a canonical direct-sum decomposition of $H^n$. For unbounded filtrations additional completeness/convergence conditions are necessary; boundedness removes those issues here.

The bounded filtered-complex convergence theorem states the following. If $F^{p}C$ is a decreasing filtration by cochain subcomplexes, preserved by the differential, finite in each cochain degree (uniform bounds $F^{a}C=C$, $F^{b}C=0$ are sufficient), then there is a spectral sequenceThe abutment filtration is $F^p{H}^n(C)=\mathrm{im}[H^n(F^{p}C)\to H^n(C)]$. Its finite length ensures stabilization and the displayed limiting-page identification. The filtered cochain complex need not itself be bounded in cochain degree. In the degreewise finite version, the bounds in degrees $n-1,n,n+1$ suffice to stabilize the terms of total degree $n$.

For a double cochain complex $D^{p,q}$ bounded in both indices, form the total cochain complex ${\mathrm{Tot}}^{n}D={⨁}_{p+q=n}{D}^{p,q}$ with anticommuting differentials $d_h,d_v$ and total differential $d_h+d_v$. If the original differentials commute, inserting the usual sign in one of them gives this convention. Filtering by the first index gives $E_1^{p,q}=H_v^q(D^{p,\ast })$ and then $E_2^{p,q}=H_h^p(H_v^{q}D)$. Filtering by the second index gives the spectral sequence taking horizontal cohomology first and vertical cohomology next. Both filtrations are finite, so both converge to $H^{\ast }(\mathrm{Tot}D)$. This is the two spectral sequences of a bounded double complex construction.

The printed left/left tensor expression needs a handedness repair. Over an arbitrary ring, a tensor product of modules over $R$ pairs a right module with a left module. To retain the order of the printed formula, take $P$ to be a bounded cochain complex of projective right $R$-modules and $M$ a left $R$-module. Alternatively, keep $P$ left, take $M$ right, and write $M{\otimes }_{R}P$ and ${\mathrm{Tor}}_{-p}^R(M,H^{q}P)$. For a commutative ring no repair is needed. We prove the first, correctly typed formulation.

Finite projective dimension gives a finite projective resolution $Q_{\bullet }\to M$ by left modules. Form $D^{p,q}=P^q{\otimes }_R{Q}_{-p}$ for $-d\le p\le 0$. On $x\otimes y$ take $d_v=d_P\otimes 1$ and $d_h=(-1{)}^{q}1\otimes d_Q$. These anticommute, and the double cochain complex is bounded in both directions. Projective modules are flat modules, so taking vertical cohomology first givesThe first identification uses flatness of $Q_{-p}$; the second is the Tor functor computed by resolving its left-module argument $M$. Balancedness of the Tor functor allows either correctly sided projective resolution to compute it, by the double-resolution argument.

Taking horizontal cohomology first instead uses flatness of $P^q$. The resolution $Q_{\bullet }$ then leaves only $P^q{\otimes }_{R}M$ in column $p=0$. Its remaining differential is $d_P\otimes 1$, so the augmentation $\mathrm{Tot}D\to P{\otimes }_{R}M$ is a quasi-isomorphism. Apply the convergence theorem to this underlying abelian-group double complex. The two spectral sequences therefore prove the Künneth spectral sequence:These are abelian groups in general; an extra module structure requires appropriate bimodule hypotheses. The index $p$ is nonpositive, so $-p$ is the nonnegative homological index of the Tor functor.

Finally write the cycle modules as $Z^q=\ker d^q$ and the boundary modules as $B^q=\mathrm{im}{d}^{q-1}$. If every boundary module $B^q$ is projective, the short exact sequence $0\to Z^q\to P^q\to B^{q+1}\to 0$ splits. Thus $Z^q$ is projective. The short exact sequence $0\to B^q\to Z^q\to H^{q}P\to 0$ is now a length-one projective resolution, so ${\mathrm{Tor}}_i^R(H^{q}P,M)=0$ for $i\ge 2$. The $E_2$ page occupies only columns $p=-1,0$. Every $d_r$ for $r\ge 2$ goes $r$ columns to the right, hence has zero source or zero target. ThereforeThis two-column degeneration of a Künneth spectral sequence gives the edge short exact sequencesDegeneration itself does not supply a canonical splitting. Projective boundaries also do not force $H^{q}P$ to be projective: a complex with differential multiplication by $2$ over $\mathbb{Z}$ has projective boundary but a $\mathbb{Z}/2\mathbb{Z}$ cohomology group.

Use homological grading $d_n:C_n\to C_{n-1}$, with zero terms below degree zero. In the projective model structure on nonnegative chain complexes, the classes are:

Thus a cofibrant complex has a projective module in each degree, and every object is fibrant. An projective acyclic fibration is equivalently a degreewise surjective quasi-isomorphism, including degree zero. To see the degree-zero assertion, lift a degree-zero homology class using the homology isomorphism, then lift the boundary discrepancy using surjectivity in degree one. Conversely a degreewise surjective quasi-isomorphism has the requisite positive-degree surjections.

Let $S^n(R)$ be the sphere chain complex with $R$ in degree $n$ and zero elsewhere. Let $D^n(R)$, for $n\ge 1$, be the disk chain complex with $R$ in degrees $n,n-1$ and identity differential. The generating cofibrations for nonnegative chain complexes areEach is a model cofibration: its degreewise cokernel is a copy of $R$ in one degree. The extra degree-zero generator must not be omitted.

Here is the lifting characterization of an acyclic chain-complex fibration. A right lifting property for $S^{n-1}(R)\to D^n(R)$ says that whenever $x\in Z_{n-1}X$ and $y\in Y_n$ satisfy $f(x)=dy$, there is $z\in X_n$ with $dz=x$ and $f(z)=y$. The generator $0\to S^0(R)$ says that $f_0$ is surjective. For a degreewise surjective quasi-isomorphism, its kernel is acyclic by the long exact sequence in homology. Choose a preliminary lift $z_0$ of $y$. Then $d{z}_0-x$ is a cycle in the kernel; subtract an element of the kernel whose boundary is this cycle. This gives the required $z$.

Conversely the lifting conditions make the kernel acyclic: take $y=0$ and any kernel cycle $x$. They also imply degreewise surjectivity inductively. Having surjectivity below degree $n$, lift $dy$ to an element of $X_{n-1}$, correct its boundary within the acyclic kernel to make it a cycle, and apply the lifting condition to lift $y$. For $n=1$ the chosen degree-zero element is already a cycle. A degreewise surjective map with acyclic kernel is a quasi-isomorphism by the long exact sequence in homology. Hence $I$-injectives are exactly acyclic fibrations.

An object $A$ is sequentially small if for every sequential diagram $Z_0\to Z_1\to \cdots$, the natural mapis bijective. Surjectivity says that a map out of $A$ factors through a finite stage; injectivity says that two such maps agreeing in the colimit agree at a later finite stage. The relative version restricts to the stated class of sequential diagrams. The domains in $I$ have this property: maps from $S^{n-1}(R)$ are cycles in one degree, and filtered colimits commute with these finite equations. Their sequential smallness is also permitted by the hint.

Apply the small object argument to the map $a:X\to Y$. Put $Z_0=X$, $p_0=a$. Given $p_k:Z_k\to Y$, take the set of all commutative squares with a generator $i:A_i\to B_i$ on the left and $p_k$ on the right. Form the pushout of the coproduct of all these $i$ along their maps into $Z_k$, obtaining $Z_{k+1}$. Each bottom map $B_i\to Y$ induces the compatible map $p_{k+1}:Z_{k+1}\to Y$. Repeat for all $k\ge 0$, and set $Z_{\infty }={\mathrm{colim}}_k{Z}_k$. This givesConcretely, attaching $S^{n-1}(R)\to D^n(R)$ adds a free generator in degree $n$ whose boundary is the chosen existing cycle; attaching $0\to S^0(R)$ adds a free degree-zero generator. Thus $j$ is an injection, and its cokernel in each degree is free on the newly attached generators. It is therefore a model cofibration; this also illustrates the relative cell complex description of the left factor.

For a lifting square into $Z_{\infty }\to Y$, sequential smallness of its domain $A_i$ factors the top map through some $Z_k$. The commutativity is already an equality in $Y$, so this is one of the squares attached at stage $k+1$. Its new cell supplies a lift $B_i\to Z_{k+1}\to Z_{\infty }$. Hence $p$ has the right lifting property with respect to every element of $I$, and is an projective acyclic fibration. Every map has the required cofibration–acyclic-fibration factorization. Only the domains need sequential smallness; there is no requirement that $X$, $Y$ or the coproduct of all cells be small.

Let $f$ be a chain map between degreewise finitely generated free abelian groups. If its reduction modulo every prime number is a quasi-isomorphism, the mapping cone acyclicity criterion makes $M(f)\otimes {\mathbb{F}}_p$ acyclic. The short exact sequence of chain complexes $0\to M\stackrel{p}{\to }M\to M\otimes {\mathbb{F}}_p\to 0$ implies that multiplication by $p$ is an isomorphism on every $H_i(M)$. These are finitely generated abelian groups. Their free summands would make multiplication by $p$ fail to be surjective, and any nonzero finite summand would make it fail to be injective for a prime dividing its order. Thus $H_i(M)=0$ and $f$ is a quasi-isomorphism. Degreewise finite generation is essential to this argument.

The model category of nonnegative complexes of left modules has quasi-isomorphisms as model weak equivalences, positive-degree surjections as model fibrations, and degreewise injections with projective cokernel as model cofibrations. All objects are fibrant; cofibrant objects are degreewise projective.

The quotient ${\mathbb{CP}}^5/{\mathbb{CP}}^2$ is five-connected and has cohomology only in degrees $0,6,8,10$, with positive products zero. Its minimal model therefore starts with closed $a_6,b_8,c_{10}$; no decomposable differential can occur below degree twelve. Sending these three generators to the cohomology basis and all later generators to zero is a chain map and a quasi-isomorphism. Thus its full minimal model is the same as that of $S^6\vee S^8\vee S^{10}$. It starts with generators $u_{11},v_{13},w_{15},t_{15}$ satisfying $du=a^2,dv=ab,dw=ac,dt=b^2$, followed by generators killing remaining products and syzygies.

For central elements, each multiplication map on the preceding quotient must be injective; a usual convention also requires the final quotient to be nonzero. The augmented Koszul complex on central ring elements tensored with $M$ is then a quasi-isomorphism to the final quotient in degree zero.

A free graded-commutative rational differential graded algebra with decomposable differentials and a quasi-isomorphism to the rational polynomial forms of a space. For simply connected finite-type spaces, $V^i$ is dual to ${\pi }_i(X)\otimes \mathbb{Q}$. A proposed finite model must have its cohomology and representing quasi-isomorphism checked; matching a list of relations alone is insufficient.

For simply connected rational spaces of finite type, isomorphic minimal Sullivan models are equivalent to the same rational homotopy type. A quasi-isomorphism between minimal models is an isomorphism, so a zigzag of cochain quasi-isomorphisms produces the same minimal model. Existence and uniqueness come from adjoining closed generators to represent missing cohomology and generators with prescribed decomposable differential to kill kernel classes, degree by degree.

The two quadratic relations form a regular sequence in $\mathbb{Q}[a,b]$, so their Koszul complex has only the quotient-ring cohomology. Represent the two degree-two classes by rational polynomial forms and choose primitives for the exact relations to obtain a quasi-isomorphism. There are no further generators.

If $X$ is simply connected and $h_{\ast }:{\pi }_{\ast }(X)\otimes \mathbb{Q}\to {\tilde{H}}_{\ast }(X;\mathbb{Q})$ is onto, choose a homogeneous homology basis and representing sphere maps into $X_{\mathbb{Q}}$. Their wedge sum gives a rational homology equivalence, hence a rational homotopy equivalence by the rational Whitehead theorem. All positive cup products therefore vanish. In a minimal model, the induced map from cohomology to the indecomposable quotient is injective; replacing a set of generators by cocycle representatives produces a quasi-isomorphism to the square-zero cohomology algebra, proving formality of a topological space.