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Use shifted degreesIts graded Lie bracket is the sign-adjusted Whitehead bracket from Question 2, equivalently the Samelson product on loops. This regrading is essential: the geometric bracket has degree minus one in the unshifted sphere grading. A free graded Lie algebra on a graded vector space $V$ means $\mathbb{L}(V)$ with the universal extension property for linear maps from $V$ to any graded Lie algebra.

Here are the general results used. The Bott–Samelson theorem says that, for a connected based CW complex $X$, the inclusion $X\to \Omega \Sigma X$ induces an isomorphism of graded algebraswith the Pontryagin product. The Milnor–Moore theorem says that for simply connected $Y$, rational loop Hurewicz homomorphism identifies $L_Y$ with the primitives of that loop-homology Hopf algebra and induces $U(L_Y)\cong H_{\ast }(\Omega Y;\mathbb{Q})$. The graded Poincaré-Birkhoff-Witt theorem identifies the associated graded algebra of $U(L)$, for its word-length filtration, with the graded symmetric algebra on $L$; hence its primitive elements of a Hopf algebra are exactly $L$ over $\mathbb{Q}$. Finally the rational Whitehead theorem identifies rational homology equivalences of simply connected CW complexes with rational homotopy equivalences. Rationalization of a topological space induces the corresponding isomorphisms of loop-space homotopy and rational homology.

Suppose first $Y$ has the rational homotopy type of $\Sigma X$. A reduced suspension is simply connected when $X$ is connected; a disconnected $X$ has $H_1(\Sigma X;\mathbb{Q})={\tilde{H}}_0(X;\mathbb{Q})\ne 0$, so it cannot have the rational homology of the simply connected $Y$. Put $V={\tilde{H}}_{\ast }(X;\mathbb{Q})$, in strictly positive degrees. Bott–Samelson theorem gives $H_{\ast }(\Omega Y;\mathbb{Q})\cong T(V)$ as an algebra. To deduce freeness of the primitive Lie algebra, its coproduct must also be considered: the initial Bott–Samelson theorem generators can have reduced-diagonal terms when $X$ has nontrivial cup products.

By Milnor–Moore theorem write this Hopf algebra as $U(L_Y)$. Its algebra indecomposables satisfyThe first identity follows from the enveloping relation: after all positive products are killed, every bracket is killed and no additional linear relation is introduced. Hence each homogeneous basis vector of $V$ has a primitive representative in $H_{\ast }(\Omega Y;\mathbb{Q})$. Choose such representatives $p_v$. Sending $v$ to $p_v$ defines an algebra map from $T(V)$ to itself whose effect on indecomposables is the identity. It is an isomorphism: $p_v=v+$ decomposable terms, and every factor in those terms has smaller positive degree than $v$. Degree induction solves recursively for each original generator and constructs the inverse. This works even with arbitrarily many generators in a degree.

Equip the source tensor algebra with primitive generators. The map is now a Hopf-algebra isomorphism, because the chosen images are primitive. The primitives in this tensor Hopf algebra are exactly $\mathbb{L}(V)$: $U(\mathbb{L}(V))=T(V)$ by the two universal properties, and the graded Graded Poincaré–Birkhoff–Witt theorem gives its primitive subspace. Consequently $L_Y\cong \mathbb{L}(V)$, proving the forward direction without assuming a primitive coproduct for the original generators.

Conversely suppose $L_Y\cong \mathbb{L}(W)$ for a positively graded $W$. Choose a homogeneous basis $w_{\lambda }$ of $W$. Represent its corresponding homotopy class by a based sphere map $S^{|w_{\lambda }|+1}\to Y_{\mathbb{Q}}$, and combine these maps intoThe wedge sum is a reduced suspension, namely $Z=\Sigma ({\bigvee }_{\lambda }{S}^{|w_{\lambda }|})$. The preceding direction, applied to this wedge sum, gives its free rational homotopy Lie algebra on the sphere inclusions. Indeed these particular tensor generators are primitive because the reduced diagonal of each sphere is zero, and loop Hurewicz homomorphism sends the adjoint sphere inclusion to that generator. Naturality of the bracket shows that $F_{\ast }$ sends the free generating basis to the prescribed free generating basis of $L_Y$. It is therefore an isomorphism of free graded Lie algebras, hence an isomorphism in every rational homotopy degree. The rational Whitehead theorem makes $F$ a rational homotopy equivalence. This proves the reverse implication and also shows that a wedge sum of spheres can be chosen as the reduced suspension.

For every chosen connected $X$ with $Y{\simeq }_{\mathbb{Q}}\Sigma X$, the required algebra statement is precisely the Bott–Samelson theorem isomorphism, transported along rational homotopy equivalence:Multiplication is concatenation of tensors and the degree-zero tensor is the unit. All freeness assertions here are graded, with the Koszul signs described above.

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

A map between simply connected spaces is a rational homotopy equivalence when it induces isomorphisms ${\pi }_n(X)\otimes \mathbb{Q}\to {\pi }_n(Y)\otimes \mathbb{Q}$ in every degree. Equivalently it induces an equivalence of their rationalizations, or a rational homology isomorphism by the rational Whitehead theorem. Spaces have the same rational homotopy type when they are connected by a zigzag of such maps. This is distinct from rational homotopy equivalence of algebraic cycles.

Study homotopy after killing torsion, especially for simply connected spaces of finite type. The rational Whitehead theorem detects rational homotopy equivalences by rational homology, while Sullivan minimal models encode rational homotopy generators and cohomological relations.

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.