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Kevin H. Lin
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darij grinberg
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When one wants to prove that a morphism $f_*$ between two chain complexes $\left(C_*\right)$ and $\left(D_*\right)$ is an isomorphismzero in homology, one of the standard approaches is to look for a chain homotopy, i. e., for a map $U_n:C_n\to D_{n+1}$ defined for every $n$ that satisfies $f_n=d_{n+1}U_n+U_{n-1}d_n$ for every $n$. However, this is not strictly necessary: For example, it is often enough to have two maps $U_n:C_n\to D_{n+1}$ and $V_n:C_n\to D_{n+1}$ defined for every $n$ that satisfy $f_n=d_{n+1}U_n+V_{n-1}d_n$ for every $n$. This way, when constructing $U_n$ and $V_n$, one doesn't have to care for them to "fit together", because each is used only one time.

However, at least my experience suggests that one does not gain much from this - when one tries to construct these $U_n$ and $V_n$, they turn out (after some simplification) to be the same.

My question is: What is the deeper reason behind this? Why do chain homotopies like to "fit together" although they don't need to?

I am sorry if this makes no sense...

EDIT: Thanks David, it seems I can't write a single absatz without a stupid mistake.

When one wants to prove that a morphism $f_*$ between two chain complexes $\left(C_*\right)$ and $\left(D_*\right)$ is an isomorphism in homology, one of the standard approaches is to look for a chain homotopy, i. e., for a map $U_n:C_n\to D_{n+1}$ defined for every $n$ that satisfies $f_n=d_{n+1}U_n+U_{n-1}d_n$ for every $n$. However, this is not strictly necessary: For example, it is often enough to have two maps $U_n:C_n\to D_{n+1}$ and $V_n:C_n\to D_{n+1}$ defined for every $n$ that satisfy $f_n=d_{n+1}U_n+V_{n-1}d_n$ for every $n$. This way, when constructing $U_n$ and $V_n$, one doesn't have to care for them to "fit together", because each is used only one time.

However, at least my experience suggests that one does not gain much from this - when one tries to construct these $U_n$ and $V_n$, they turn out (after some simplification) to be the same.

My question is: What is the deeper reason behind this? Why do chain homotopies like to "fit together" although they don't need to?

I am sorry if this makes no sense...

When one wants to prove that a morphism $f_*$ between two chain complexes $\left(C_*\right)$ and $\left(D_*\right)$ is zero in homology, one of the standard approaches is to look for a chain homotopy, i. e., for a map $U_n:C_n\to D_{n+1}$ defined for every $n$ that satisfies $f_n=d_{n+1}U_n+U_{n-1}d_n$ for every $n$. However, this is not strictly necessary: For example, it is often enough to have two maps $U_n:C_n\to D_{n+1}$ and $V_n:C_n\to D_{n+1}$ defined for every $n$ that satisfy $f_n=d_{n+1}U_n+V_{n-1}d_n$ for every $n$. This way, when constructing $U_n$ and $V_n$, one doesn't have to care for them to "fit together", because each is used only one time.

However, at least my experience suggests that one does not gain much from this - when one tries to construct these $U_n$ and $V_n$, they turn out (after some simplification) to be the same.

My question is: What is the deeper reason behind this? Why do chain homotopies like to "fit together" although they don't need to?

I am sorry if this makes no sense...

EDIT: Thanks David, it seems I can't write a single absatz without a stupid mistake.

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darij grinberg
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Chain homotopy: Why du+ud and not du+vd?

When one wants to prove that a morphism $f_*$ between two chain complexes $\left(C_*\right)$ and $\left(D_*\right)$ is an isomorphism in homology, one of the standard approaches is to look for a chain homotopy, i. e., for a map $U_n:C_n\to D_{n+1}$ defined for every $n$ that satisfies $f_n=d_{n+1}U_n+U_{n-1}d_n$ for every $n$. However, this is not strictly necessary: For example, it is often enough to have two maps $U_n:C_n\to D_{n+1}$ and $V_n:C_n\to D_{n+1}$ defined for every $n$ that satisfy $f_n=d_{n+1}U_n+V_{n-1}d_n$ for every $n$. This way, when constructing $U_n$ and $V_n$, one doesn't have to care for them to "fit together", because each is used only one time.

However, at least my experience suggests that one does not gain much from this - when one tries to construct these $U_n$ and $V_n$, they turn out (after some simplification) to be the same.

My question is: What is the deeper reason behind this? Why do chain homotopies like to "fit together" although they don't need to?

I am sorry if this makes no sense...