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Dmitri Pavlov
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Consider the simplicial set $\def\S{{\bf S}} \def\Sing{\mathop{\rm Sing}} \S^n=Δ^n/∂Δ^n$, which has exactly two nondegenerate simplices: a 0-simplex and an $n$-simplex.

Consider the map $\S^n→\Sing S^n$ that sends the only vertex of $\S^n$ to the given basepoint of $S^n$ and the only nondegenerate $n$-simplex of $\S^n$ to some singular $n$-simplex $Δ^n→S^n$ that sends the boundary to the basepoint and induces a degree 1 map once we mod out the boundary.

The map $\S^n→\Sing S^n$ is a simplicial weak equivalence. Thus, the induced map on integral normalized simplicial cochains $\def\Z{{\bf Z}} \def\C{{\rm C}} \C(S^n,\Z)=\C(\Sing S^n,\Z)→\C(\S^n,\Z)$ is a quasi-isomorphism of augmented differential graded rings.

Observe that $\C(\S^n,\Z)$ is a differential graded ring with exactly two generators: one in degree 0 and another one in degree $n$. The differentials of both generators vanish. Furthermore, the degree $n$ generator squares to 0 for dimensional reasons. Thus, the normalized simplicial cochain ring $\C(\S^n,\Z)$ is precisely the graded cohomology ring $\def\H{{\rm H}} \H(S^n,\Z)$.

Consider the simplicial set $\def\S{{\bf S}} \def\Sing{\mathop{\rm Sing}} \S^n=Δ^n/∂Δ^n$, which has exactly two nondegenerate simplices: a 0-simplex and an $n$-simplex.

Consider the map $\S^n→\Sing S^n$ that sends the only vertex of $\S^n$ to the given basepoint of $S^n$ and the only nondegenerate $n$-simplex of $\S^n$ to some singular $n$-simplex $Δ^n→S^n$ that sends the boundary to the basepoint and induces a degree 1 map once we mod out the boundary.

The map $\S^n→\Sing S^n$ is a simplicial weak equivalence. Thus, the induced map on integral normalized simplicial cochains $\def\Z{{\bf Z}} \def\C{{\rm C}} \C(S^n,\Z)=\C(\Sing S^n,\Z)→\C(\S^n,\Z)$ is a quasi-isomorphism of augmented differential graded rings.

Observe that $\C(\S^n,\Z)$ is a differential graded ring with exactly two generators: one in degree 0 and another one in degree $n$. The differentials of both generators vanish. Furthermore, the degree $n$ generator squares to 0 for dimensional reasons. Thus, $\C(\S^n,\Z)$ is precisely the graded ring $\def\H{{\rm H}} \H(S^n,\Z)$.

Consider the simplicial set $\def\S{{\bf S}} \def\Sing{\mathop{\rm Sing}} \S^n=Δ^n/∂Δ^n$, which has exactly two nondegenerate simplices: a 0-simplex and an $n$-simplex.

Consider the map $\S^n→\Sing S^n$ that sends the only vertex of $\S^n$ to the given basepoint of $S^n$ and the only nondegenerate $n$-simplex of $\S^n$ to some singular $n$-simplex $Δ^n→S^n$ that sends the boundary to the basepoint and induces a degree 1 map once we mod out the boundary.

The map $\S^n→\Sing S^n$ is a simplicial weak equivalence. Thus, the induced map on integral normalized simplicial cochains $\def\Z{{\bf Z}} \def\C{{\rm C}} \C(S^n,\Z)=\C(\Sing S^n,\Z)→\C(\S^n,\Z)$ is a quasi-isomorphism of augmented differential graded rings.

Observe that $\C(\S^n,\Z)$ is a differential graded ring with exactly two generators: one in degree 0 and another one in degree $n$. The differentials of both generators vanish. Furthermore, the degree $n$ generator squares to 0 for dimensional reasons. Thus, the normalized simplicial cochain ring $\C(\S^n,\Z)$ is precisely the graded cohomology ring $\def\H{{\rm H}} \H(S^n,\Z)$.

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Dmitri Pavlov
  • 37.8k
  • 4
  • 97
  • 183

Consider the simplicial set $\def\S{{\bf S}} \def\Sing{\mathop{\rm Sing}} \S^n=Δ^n/∂Δ^n$, which has exactly two nondegenerate simplices: a 0-simplex and an $n$-simplex.

Consider the map $\S^n→\Sing S^n$ that sends the only vertex of $\S^n$ to the given basepoint of $S^n$ and the only nondegenerate $n$-simplex of $\S^n$ to some singular $n$-simplex $Δ^n→S^n$ that sends the boundary to the basepoint and induces a degree 1 map once we mod out the boundary.

The map $\S^n→\Sing S^n$ is a simplicial weak equivalence. Thus, the induced map on integral normalized simplicial cochains $\def\Z{{\bf Z}} \def\C{{\rm C}} \C(S^n,\Z)=\C(\Sing S^n,\Z)→\C(\S^n,\Z)$ is a quasi-isomorphism of augmented differential graded rings.

Observe that $\C(\S^n,\Z)$ is a differential graded ring with exactly two generators: one in degree 0 and another one in degree $n$. The differentials of both generators vanish. Furthermore, the degree $n$ generator squares to 0 for dimensional reasons. Thus, $\C(\S^n,\Z)$ is precisely the graded ring $\def\H{{\rm H}} \H(S^n,\Z)$.