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Given two morphisms between chain complexes $f_\bullet,g_\bullet\colon\,C_\bullet\longrightarrow D_\bullet$, a chain homotopy between them is a sequence of maps $\psi_n\colon\,C_n\longrightarrow D_{n+1}$ such that $f_n-g_n= \partial_D \psi_n+\psi_{n-1}\partial_C$. I can motivate this definition only when the chain complexes are associated to some topological space. For example if $C_\bullet$ and $D_\bullet$ are simplicial chain complexes, then for a simplex $\sigma$, a homotopy between $f(\sigma)$ and $g(\sigma)$ is something like $\psi(\sigma)\approx\sigma\times [0,1]$, whose boundary is $f(\sigma)-g(\sigma)-\psi(\partial\sigma)$.
But chain homotopy also features in contexts where there is no topological space lurking in the background. Examples are chain homotopy of complexes of graphs (e.g. Conant-Schneiderman-Teichner) or chain homotopy of complexes in Khovanov homology. In such contexts, the motivation outlined in the previous paragraph makes no sense, with "the boundary of a cylinder being the top, bottom, and sides", because there's no such thing as a "cylinder". Thus, surely, the "cylinder motivation" isn't the most fundamental reason that chain homotopy is "the right" relation to study on chain complexes. It's an embarassing question, but:

What is the fundamental (algebraic) reason that chain homotopy is relevant when studying chain complexes?
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If you believe simplicial things are topological then the Dold-Kan theorem tells us that there is topology. –  Sean Tilson Mar 24 '11 at 0:28
Two morphisms of complexes are homotopic iff their difference factors over a split acyclic complex (i.e. spliced together from split short exact sequences). So the homotopy category of complexes is the factor category (in the sense of additive categories) of the category of complexes modulo the full additive subcategory of split acyclic complexes; i.e. $\text{K}(A) = \text{C}(A)/\text{C}_\text{split acyclic}(A)$, for an additive category $A$. –  Matthias Künzer Mar 24 '11 at 8:36
@Sean Thanks for pointing that out! Are you saying that I would be getting an "honest cylinder" on the simplicial side of the Dold-Kan correspondence? Is so, how would the cylinder make sense in terms of the algebraic objects which I started out with? –  Daniel Moskovich Mar 24 '11 at 15:15
@Daniel: One choice of cylinder is the chain complex associated with the free abelian group on the $1$-simplex. The Dold-kan map is an equivalence of categories, but it is not a monoidal functor. The theorem of Eilenberg-Zilber tells us that it is monoidal up to homotopy (so the cylinder $S\otimes_{\Delta} I\mapsto N(C)\otimes_{\mathrm{chain}} N(C)$. If I remember correctly, though, there is a simpler choice of cylinder on the chain complex side. –  Harry Gindi Mar 25 '11 at 1:45

6 Answers 6

up vote 21 down vote accepted

Here's one way to look at it: There is a chain complex $Hom(C,D)$ in which the $n$th chain group is the product over $k$ of $Hom(C_k,D_{n+k})$ and the boundary is given by $\partial (f(c))=(\partial f)(c)+(-1)^{|f|}f(\partial c)$. A chain map is a $0$-cycle, and two of them are chain homotopic if they differ by a boundary.

EDIT: And then a chain map $B\to Hom(C,D)$ corresponds precisely to a chain map $B\otimes C\to D$, where $B\otimes C$ is defined using the usual convention $\partial (b\otimes c)=(\partial b)\otimes c + (-1)^{|b|}b\otimes \partial c$

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To spell it out in connection with the question: If $I$ denotes the cellular chain complex of the interval, then a chain map $I\to \text{hom}(C,D)$ is adjoint to a chain map $I\otimes C \to D$ which in turn is equivalent to specifying a chain homotopy between $0 \otimes C \to D$ and $1 \otimes C \to D$. –  John Klein Mar 24 '11 at 2:03
@Tom: is it just me, or should the last term be the boundary be $-(-1)^{|f|} f(\partial c)$ as in Alan Wilder's answer? If $|f| = 1$ you don't get the right definition of chain homotopy, and if $|f| = 0$ you don't get the right definition of a chain map. –  Qiaochu Yuan May 19 '11 at 20:21
@Qiaochu: If $|f|=0$ then my equation says that $\partial f=0$ means $f\circ\partial=\partial\circ f$. If $f=1$ then my equation says that $\partial\circ f+f\circ\partial$ is a boundary. –  Tom Goodwillie May 19 '11 at 21:34
@Tom: oh, I see. I mixed up $\partial(f(c))$ and $(\partial f)(c)$ in your notation. It seems to me more natural to have $(\partial f)(c)$ on the LHS...? –  Qiaochu Yuan May 20 '11 at 14:57
@Qiaochu: I did write it backwards, from the point of view of defining $\partial f$. The reason I wrote it that way was to make the sign look inevitable. I mean, using the rule of thumb that $\partial B(x,y)$ is $B(\partial x,y)+(-1)^{|x|}B(x,\partial y)$ if $B$ is a bilinear expression. (Unless you have to worry about $\partial B$ itself, in which case there's another term $(\partial B)(x,y)$. On the right, that is.) –  Tom Goodwillie May 20 '11 at 15:55

There is an inner-hom in $\mathbf{Chain}$, and the 1-chains are chain homotopies. The definition is $$ \underline{\mathbf{Chain}}(C_\bullet,D_\bullet)_k = \Pi_n \textrm{Hom} (C_{n-k},D_n) $$ so a 0-chain is just a map $f_n:C_n\rightarrow D_n$. The differential of this complex is given by $$ df(c) = d_D(f(c)) - (-1)^{|f|}\left(f(d_C(c)\right) $$ So a 0-cycle is just a chain map. A 1-chain whose boundary is $f-g$ is exactly a chain homotopy from $g$ to $f$.

Alternatively, there is a model structure on $\mathbf{Chain}$ where the weak equivalences are quasi-isomorphisms, and you can make sense of cylinder as a cylinder object for a chain complex, and then your topological motivation should all still make sense. I don't know all the details though so I won't try...

EDIT: crosspost...

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This is mostly just to expand a little on John Klein and Alan Wilder's statements.

If you take the category of chain complexes and formally degree that quasi-isomorphisms $C \to D$ should become isomorphisms, you are already forced to identify chain-homotopic maps together.

Let $I$ be the chain complex which is $\mathbb{Z} \cdot s \oplus \mathbb{Z} \cdot t$ in degree 0, and $\mathbb{Z} \cdot H$ in degree 1, with $\partial H = t - s$. Then for any complex $C$, there are quasi-isomorphisms $i_s(c) = s \otimes c$ and $i_t(c) = t \otimes c$ from $C$ to $I \otimes C$, and an inverse quasi-isomorphism $p: I \otimes C \to C$ which kills $H$ and sends $s,t$ to $1$. We have $p i_s = id = p i_t$, and so if we turn quasi-isomorphisms into isomorphisms we get the identification $i_s = p^{-1} = i_t$.

A chain homotopy $H$ between two maps $f,g:C \to D$ is the same as a map $h: I \otimes C \to D$ such that $h i_s = f$, $h i_t = g$, with $h(H \otimes c) = H(c)$. If we have decreed quasi-isomorphisms to be isomorphisms, then we find $f = h i_s = h i_t = g$.

The miracle is that, for nice complexes, chain homotopy equivalence is enough. This is where the analogy with topological spaces comes in - the complex $I \otimes C$ is a cylinder object in a suitable model structure.

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Here is how I think about chain complexes. (I mention that you should always allow yourself complexes that go in both directions.)

There is a category of $\mathbb Z$-graded vector spaces, which is monoidal (just because it's the category of $\mathbb G_m$-modules), but I give it the interesting symmetric structure following the usual "super" sign rules --- let $\mathfrak Q$ denote the one-dimensional vector space in degree $1$; then I choose the braiding $\mathfrak Q \otimes \mathfrak Q \to \mathfrak Q \otimes \mathfrak Q$ to be minus identity (this determines the braiding on the category).

Now, give the object $\mathfrak Q$ the structure of an abelian Lie algebra. (Note that by "Lie algebra" and "representation" I mean that you should interpret all AS/IHX/STU with respect to the chosen braiding.)

Then the category of Chain Complexes, as a symmetric monoidal category, is precisely the category of (left, say) $\mathfrak Q$-modules: checking this is a nice exercise (and it explains what are the correct signs).

But the category of $\mathfrak Q$-modules, as Alan Wilder has said, has an inner hom, and you can tell what it is because it's an inner hom of representations of a Lie algebra. Namely, if $f\in \underline{\hom}(X,Y)$, then the basis vector $d\in \mathfrak Q$ acts on $f$ by $f \mapsto [d,f]$, where the commutator is, of course, to be interpreted in the super sense, which is the sense internal to the category.

So chain maps are the invariant global (= degree-0) elements. (degree-(-1)) $h$ is a homotopy between (degree-0) $f,g$ iff $[d,h] = f-g$, as Tom pointed out. Another way to say this: $f-g$ is in the $\mathfrak Q$-orbit generated by the degree-(-1) elements, so that $f,g$ are in the same coset thereof.

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It is simpler, I think, to say that the category of graded vector spaces is monoidal because it is the catergory of $k\mathbb Z$-comodules. Potatos potatoes, I guess... –  Mariano Suárez-Alvarez Mar 24 '11 at 3:09
Could you please expand on what "AS/IHX/STU" means? –  Peter McNamara Mar 24 '11 at 3:43
@Mariano: Sure. But I find that I often talk to algebraically-minded people, who prefer names like "$\mathbb G_m$". –  Theo Johnson-Freyd Mar 24 '11 at 15:02
@Peter: Since I happened to be writing back to Daniel Moskovich, who happens to be an expert in finite-type knot invariants, I used a standard shorthand. I mean only the following. One can present the notion of "Lie algebra" and "module of a Lie algebra" in terms of some operad (which must be "typed" for the notion of module); recall that an operad is (equivalent data to) a symmetric monoidal category with some axioms, and that an "algebra for an operad" is a symmetric monoidal functor from that operad to your chosen category. The AS/IHX/STU relations are generators for the operad. –  Theo Johnson-Freyd Mar 24 '11 at 15:06
@Peter: I realize that I might have made it only more opaque. Here is is in the particular case at hand. A lie algebra in a symmetric monoidal category $(\mathcal C,\otimes,\text{flip})$ is an object $L$ and a "bracket" $\text{bracket} : L\otimes L \to L$. The AS relation says that $\text{bracket}\circ\text{flip} = -\text{bracket}$. The IHX relation is the Jacobi axiom: $\text{bracket}\circ(\text{bracket}\otimes\text{id})\circ(\text{id}+\tau+\tau^2)‌​=0$, where $\tau$ is the cyclic permutation of $L^3$. It's called this because if you draw the bracket as a trivalent vertex (continued) –  Theo Johnson-Freyd Mar 24 '11 at 15:10

An important application of chain homotopy is in (co)homology: If $f,g: C \to D$ are homotopic then $$[f_n] = [g_n]: H_n(C) \to H_n(D).$$

Moreover chain homotopy has good functorial properties. For example, with $f,g$, the chain maps $$ Hom(f,id_B), Hom(g,id_B): Hom(D,B) \to Hom(C,B)$$ are again homotopic, implying
$$ [Hom(f_n,id_B)] = [Hom(g_n,id_B)]: H^n(D,B) \to H^n(C,B).$$ As an application, one can show that induced homomorphisms for $Ext$ are unique. For: Let $C \to A$ resp. $D \to A^'$ be a projective resolution of $A$ resp. $A^'$ and let $\alpha: A \to A^'$ be a homomorphism. By the fundamental lemma of projective resolutions, two chain maps $f, g$ that extend $\alpha$ are homotopic, so by the above, $$ \alpha^*_n := [Hom(f_n,id_B)]: Ext^n(A^',B) \to Ext^n(A,B)$$ is well-defined, where $Ext^n(A,B):= H^n(C,B)$.

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This was a comment, but I guess it was not so clear, although it is hiding in some of the above answers:

There is an object in $Ch^+(R)$ that behaves like an interval. Think of the simplicial interval, and you get a chain complex $I_*$ over $R$ with $I_0=Re_0 \oplus Re_1$, $I_1=Rf_0$ and all other groups 0, with differential $d(f_0)=e_1−e_0$. This plays the role of the unit interval. Now a chain homotopy of two chain maps $f,g:A_∗ \to B_∗$ really is $H:A_∗ \otimes I_* \to B_*$ where the tensor product takes place in R chain complexes. I am not sure how well this object behaves with respect to the Dold-Kan maps, but it certainly is the normalized chains on $\Delta^1$. You could similarly look at maps of $R$ complexes from $I_*$ into $Hom_R(A_*,B_*)$.

With this in hand you can form cylinder objects and cofiber sequences in the "obvious" way, and things start looking a bit more geometric.

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This was clear to me- but a cylinder object is not a genuine topological cylinder, so it's not clear to me how to make sense of the topological motivation in this context- exactly because I don't know how this object behaves with respect to Dold-Kan maps. But the algebraic motivation given in the answers does provide an answer to my question- thanks to everyone! –  Daniel Moskovich Mar 25 '11 at 3:50

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