Let $(C_*,d)$ be a chain complex of vector spaces over a field $F$. Can always be constructed a homotopy equivalence $C_* \rightarrow H(C_*)$ ? (Here $H(C_*)$ is seen as a chain complex with $d=0$)
There is a split exact sequence $0 \rightarrow B_* \rightarrow Z_* \rightarrow H_*(C) \rightarrow 0$ so we have a map $H_*(C) \rightarrow Z_* \subseteq C_*$ which induces a quasi-isomorphism. Is this map the homotopy equivalence between $C_*$ and its homology?
Best Answer
In the case of a field, this is true. We can write $Z = B\oplus i(H)$ for $i : H\to Z$ a section of the projection and $C = L\oplus B\oplus i(H)$. Define $j : H\to C$ using $i$, and define $q : C\to H$ by sending $L$ and $B$ to $0$, and projecting $i(H)$ back.
These are maps of complexes: $qd = 0 = dq$ since $H(C)$ has trivial differential, and $jd=dj=0$ since $j$ lands on cycles. Moreover, by construction, $qi = 1$.
Finally define $h : C\to C$ by picking a section $s : B\to C$ of $d : C\to B$ and sending $b\in B$ to $s(b)$, and $L$ and $i(H)$ to $0$. Then,
so $dh+hd = 1 - qi$ is a homotopy between $1$ and $iq$.