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The homotopy fibre is the homotopy pullback of the path fibration over the target sphere along $f$. The Sullivan fibre-model theorem computes it by taking a relative acyclic model of that path fibration and tensoring it over the target model with the source model. Here the construction can be written out and checked directly.

Adjoin $a$ of degree $2n-1$ and $b$ of degree $4n-2$ to the sphere model, withThis is a differential, because $D(v-au)=u^2-u^2=0$. It is an acyclic relative Sullivan algebra: putting $v^{\prime }=v-au$ makes its two pairs $Da=u$ and $Db=v^{\prime }$. The first is a polynomial generator killed by an odd generator; the second is an odd generator killed by an even polynomial generator. Each pair has cohomology only $\mathbb{Q}$, the second by the ordinary polynomial antiderivative calculation in characteristic zero. Hence this models the contractible path space, and its augmentation models the chosen basepoint.

Tensoring with the projective-space model via the map in part (i) givesThe last differential is necessary: using only $Db=f^{\ast }v$ would give $D^{2}b=x^{2n}\ne 0$. Make the invertible generator change $z=y-xa$, of degree $2n+1$. Since $x$ is even,For $n\ge 2$, all nonzero differentials are decomposable and the generator ordering $x,a,z,b$ is Sullivan. Thus no further minimization is needed:For example at $n=2$ this is $(\Lambda (x_2,a_3,z_5,b_6),Da=x^2,Db=xz)$.

The indecomposable generators of a simply connected finite-type minimal model are dual to its rational homotopy groups. Therefore, for $n\ge 2$,These four degrees are distinct. The exact homotopy sequence provides an independent check: ${\mathbb{CP}}^n$ has rational homotopy only in degrees $2,2n+1$, while $S^{2n}$ has it in degrees $2n,4n-1$. For $n\ge 2$ these source and target degrees do not overlap, so the source classes survive in the fibre and the target classes contribute their one-degree-lower connecting classes, giving the same four groups.

At $n=1$, the displayed relative fibre algebra is not minimal: $Da=x$ and $Db=z$ are linear pairs. Cancelling both acyclic pairs leaves $\mathbb{Q}$. AccordinglyThis agrees with ${\mathbb{CP}}^1\cong S^2$ and the degree-one map being a homotopy equivalence. The case distinction is required before reading homotopy groups from the generators.