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.
Best Answer
EDITED, because I think I see the big picture now.
We have the following theorem: let $C$ and $D$ be complexes of projective objects. The following are equivalent: (a) the map $f: C \to D$ is the zero map in the derived category (b) there is a homotopy between $f$ and $0$.
In this theorem, the definition of homotopy is that $f=du+ud$. So my answer to your question is: In practice, if a map induces the zero map on homology, it is probably zero in the derived category. There do exist maps which are of the form $du+vd$, but are not of the form $du+ud$, see my other answer.
Paragraph about hereditary algebras deleted because I don't think it was quite right.
Some more elementary observations
(1) It is required that $f$ be a map of chain complexes, so $df=fd$. So we want $d(du+vd) =(du+vd)d$ or $dud=dvd$. This doesn't force $u=v$, but it is the easiest way to achieve it.
(2) There is a topological way of thinking of the condition $du+ud=f$, which I learned from Joel Kamnitzer. Let $I$ be the chain complex with $I_1=\mathbb{Z}$, $I_0 = \mathbb{Z}^2$ and the map $I_1 \to I_0$ given by $(1 \ -1)$. Let $\partial I$ be the subchain complex where $(\partial I)\_0=I_0$ and $(\partial I)\_i=0$ for all other $i$.
Then writing $f=du+ud$ is equivalent to finding a map $u:C \times I \to D$ such that, when we restrict to $C \times (\partial I)$, we have the map $f$ on one component and $0$ on the other. $I$ is the chain complex of the obvious triangulation of the unit interval. Thinking of $I$ as the unit interval, this really is a homotopy between $f$ and $0$. I can't think of an analogous geometric motivation for $f=du+vd$.