# Do any 2 morphisms from objects $X$ to $Y$ define a chain homotopy equivalence?

I was curious about one thing: Let $A$ and $B$ be abelian categories with enough projectives, let $X$, $Y$ be objects in $A$ and let $P_{\bullet} \rightarrow X$, $P'_{\bullet} \rightarrow Y$ be projective resolutions of $X$ and $Y$ respectively, this already well-known theorem (http://bit.ly/1t0y2Rm, theorem 6.9), says that any morphism $f: X \rightarrow Y$ defines a chain map $f_{\bullet}: P_{\bullet} \rightarrow P'_{\bullet}$ and any two chain maps $f_{\bullet},g_{\bullet}: P_{\bullet} \rightarrow P'_{\bullet}$ are chain homotopic. So that means that any two morphisms $f,g: X \rightarrow Y$ define a chain homotopy being that they both define chain maps $f_{\bullet}$, $g_{\bullet}: P_{\bullet} \rightarrow P'_{\bullet}$?

Let's say I want to compute the derived functor of some right exact functor $F: A \rightarrow B$ for object $Y$ and resolution $P'_{\bullet} \rightarrow Y$, since chain homotopies define quasi-isomorphisms which are preserved by additive functors then if I have a resolution $P_{\bullet} \rightarrow X$ and a morphism $f: X \rightarrow Y$ I can just use the zero morphism $0_{XY}: X \rightarrow Y$ and get a chain homotopy between $f_{\bullet}$ and $0_{XY_{\bullet}}$ and there'll be a chain homotopy equivalence between $P_{\bullet}$ and $P'_{\bullet}$ and I can compute $L_nF(Y)$? So any morphism that is not trivial defines a chain homotopy since I can always use the zero morphism to define a chain homotopy? Does that mean that objects and their subobjects have the same derived functor groups for any right (left) exact functor? For any subobject $A' \subseteq A$ I can always define a chain homotopy between the chain maps induced by the monomorphism $A' \rightarrow A$ and the zero morphism to define a chain homotopy between their resolutions and compute derived functors?

## 1 Answer

It looks like you're misreading the theorem in a critical detail. It's not the case that any two maps $f_\bullet,g_\bullet:P_\bullet\to P'_\bullet$ are chain homotopic, but that any two such maps induced by the same map $f:X\to Y$ are. (For a counterexample, if $X$ and $Y$ are themselves projective, then two chain maps between $X$ and $Y$ are chain homotopic if and only if they're equal.) This shows that your procedure for computing the left derived functor won't work out. And that's good news, because if it did the whole theory would be rather trivial!

• Thank you! That's why I was confused, let me reread the proof again and I'll get back to you. – JustC Sep 4 '14 at 1:48
• This may sound dumb, but are chain homotopies something that can be used to compute derived functors? Like if I know $L_nF(X)$ and want to compute $L_nF(Y)$, so I try to find a morphism $f: X \rightarrow Y$ that induces a chain homotopy between 2 chain maps so that I can compute $L_nF(Y)$?... or were chain homotopies mainly used as foundational tools to prove things like indepence of choice of resolution? This is pure ignorance, but is there such a thing as "chain homotopy" methods to compute derived functors? I've mainly only seen them mentioned in foundational stages of homological algebra. – JustC Sep 5 '14 at 1:56
• A map induces a chain map, which induces maps of derived functor values. Two chain homotopic maps induce the same map on the derived functor values. One reason you may not see them much is that this is sufficient, but not necessary-the homotopy category is more finely structured than is appropriate for studying derived functors. – Kevin Carlson Sep 5 '14 at 2:02
• Now it's much clearer, thanks again for taking your time to answer. – JustC Sep 6 '14 at 0:10
• Glad it was helpful! – Kevin Carlson Sep 6 '14 at 0:57