Kolby Castillo
2022-09-26
Answered

Let C be a chain complex over a commutative ring R. Let M be a subchain of C such that M is chain equivalent to the zero chain complex. Must C/M and C be chain equivalent ?

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Willie Sharp

Answered 2022-09-27
Author has **8** answers

Step 1

Take complexes of abelian groups:

$$C:=\cdots \u27f60\u27f6\mathbb{Z}\stackrel{\times 2}{\u27f6}\mathbb{Z}\u27f60\u27f6\dots ,$$

$$M:=\cdots \u27f60\u27f6\mathbb{Z}\stackrel{\times 2}{\u27f6}2\mathbb{Z}\u27f60\u27f6\dots ,$$

$$C/M:=\cdots \u27f60\u27f60\u27f6\mathbb{Z}/2\mathbb{Z}\u27f60\u27f6\dots .$$

Step 2

Then M is isomorphic to the complex

$$\cdots \u27f60\u27f6\mathbb{Z}\stackrel{\text{id}}{\u27f6}\mathbb{Z}\u27f60\u27f6\dots ,$$

which is contractible (homotopy equivalent to the zero complex), but there are no nonzero maps $C/M\to C$, so C and C/M are not homotopy equivalent.

Take complexes of abelian groups:

$$C:=\cdots \u27f60\u27f6\mathbb{Z}\stackrel{\times 2}{\u27f6}\mathbb{Z}\u27f60\u27f6\dots ,$$

$$M:=\cdots \u27f60\u27f6\mathbb{Z}\stackrel{\times 2}{\u27f6}2\mathbb{Z}\u27f60\u27f6\dots ,$$

$$C/M:=\cdots \u27f60\u27f60\u27f6\mathbb{Z}/2\mathbb{Z}\u27f60\u27f6\dots .$$

Step 2

Then M is isomorphic to the complex

$$\cdots \u27f60\u27f6\mathbb{Z}\stackrel{\text{id}}{\u27f6}\mathbb{Z}\u27f60\u27f6\dots ,$$

which is contractible (homotopy equivalent to the zero complex), but there are no nonzero maps $C/M\to C$, so C and C/M are not homotopy equivalent.

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