$\DeclareMathOperator{\Ch}{\mathit{Ch}}$Let $\Ch_\mathbb{Q}$ denote the model category of chain complexes over rational numbers. Let $T_\ast$ be a tree in $\Ch_{\mathbb{Q}}$ with $n$ vertices.

How to classify trees with respect to weak equivalences i.e., chain homotopies? Is it true that the classification can be recovered from the $\mathit{ho}(\Ch_{\mathbb{Q}})$?

I think the key factor is that any chain complex $C_\ast\cong \oplus_n V\langle n\rangle_\ast$, here $V\langle n\rangle_\ast$ is the chain complex concentrated in degree $n$ and $V\langle n\rangle_n= H_n(C_\ast)$

For example if we take a path with length 2, $f_\ast : C_\ast \to C_\ast'$ then it is equivalent to maps $H_n(f_\ast): H_n(C_\ast) \to H_n(C_\ast')$, that is maps between vector spaces. We know that any map between vector spaces is completely describe by the dim(Ker). In this case, any path of length 2 is fully describe by $\dim(\mathrm{ker}(H_n(f_\ast)))_n$.

I really appreciate it if someone could say something for trees.

  • 3
    $\begingroup$ To clarify: if $T$ is a tree, a "tree in $Ch_{\mathbb Q}$" is the same thing as a functor $F(T) \to Ch_{\mathbb Q}$ where $F(T)$ is the free category on $T$ (thought of as a graph), right? In case you're not aware, the theory of quiver representations should be of interest to you. $\endgroup$
    – Tim Campion
    Oct 12, 2020 at 17:41
  • $\begingroup$ @TimCampion: Thanks. Is there any classification of the quiver representations of the form of a tree? $\endgroup$ Oct 12, 2020 at 19:04
  • $\begingroup$ Gabriel's theorem says that if $T$ is a Dynkin tree then the quiver representations of $T$ are well-behaved. I'm not sure how much is known about general trees. By the way, when you say "tree", do you just mean that the underlying undirected graph is a tree, or do you intend a condition on the orientations of the edges? $\endgroup$
    – Tim Campion
    Oct 12, 2020 at 19:20
  • 1
    $\begingroup$ There are small trees with wild representation type, for instance trees having one more vertex than affine Dynkin diagrams of type E. $\endgroup$
    – F. C.
    Oct 12, 2020 at 19:34

2 Answers 2


Is it true that the classification can be recovered from the ho(ChQ)?

Yes, the canonical functor Ho(Fun(T,M))→Fun(T,Ho(M)) induces a bijection on isomorphism classes if T is a tree and M is the relative category of rational chain complexes.

The canonical inclusion ι of the relative category of graded rational vector spaces and isomorphisms into the relative category of rational chain complexes and quasi-isomorphisms given by equipping graded rational vector spaces with a zero differential preserves weak equivalences. The functor H in the opposite direction computes homology, it also preserves weak equivalences.

These facts can be used to establish the above bijection as follows.

For surjectivity, take any functor hF: T→Ho(M) and construct F: T→M as follows. On objects, F(X)=hF(X). On morphisms, choose a lift of hF(e) along M→Ho(M) for each edge e of the tree. The values of F on other morphisms of T are now uniquely determined because any morphism of T is a composition of a unique sequence of edges. We bifibrantly replace the resulting diagram in the projective (respectively injective) model structure on functors T→M, depending on which way the tree T grows.

The diagrams F and ιHF are weakly equivalent. If T has a single object, this is established by choosing noncanonical splittings C_n = B'_n ⊕ H_n ⊕ B_n. If T has more than one object, we construct such a weak equivalence by induction on the tree, starting at the root and proceeding to the leaves. At each step, we choose the splitting C_n = B'_n ⊕ H_n ⊕ B_n so that it is compatible with the given map H_n(q)→H_n(p) (respectively H_n(p)→H_n(q)), where q is the parent of p in the tree.

For injectivity, suppose that F,G: T→M are bifibrant. We have to show that any natural isomorphism t: H(F)→H(G) can be promoted to a natural quasi-isomorphism F→G. This is done in the manner of the previous paragraph: start at the root and proceed inductively to the leaves. The transition maps F(q)→F(p) and G(q)→G(p) are injections, so we can always make a compatible choice of a quasi-isomorphism.

As pointed out in the comments, the above fact holds in a more general setting: the canonical functor Ho(Fun(T,M))→Fun(T,Ho(M)) is essentially surjective, full, and conservative for any model category M and tree T. See Proposition 2.15 and 2.20 in Catégories dérivables by Denis-Charles Cisinski.

  • $\begingroup$ I assume that $H: Ch_k \to Ho(Ch_k) = Gr_k$ is the homology functor and that $\iota : Ho(Ch_k) \to Ch_k$ is its "zero-differential" section. In the second-to-last paragraph, I don't understand where you get a map $Hom(F,G)/\sim \to Hom(\iota H(F), \iota H(G))/\sim$ from -- the quasi-isomorphisms between $F(v)$ and $\iota HF(v)$ are not natural for $v \in T$, so it's not immediately obvious that these levelwise maps can be made into natural maps between $F$ and $G$. $\endgroup$
    – Tim Campion
    Oct 12, 2020 at 19:12
  • $\begingroup$ @TimCampion: I added a new paragraph explaining this. $\endgroup$ Oct 12, 2020 at 19:26
  • 1
    $\begingroup$ Also, your second paragraph, you claim that $\iota, H$ are DK-equivalences. I don't think this is true -- $Gr_k$ is a 1-category, but the DK-localization of $Ch_k$ has higher homotopy in its hom-spaces (given by degree-shifting maps). $\endgroup$
    – Tim Campion
    Oct 12, 2020 at 20:05
  • 1
    $\begingroup$ More explicitly, cofibrantly replace $F$ by $F' = (k[n] \to D[n+1])$. Then $Hom(F',G)$ is 1-dimensional and $Hom(\Sigma F', G) = 0$, so the only nullhomotopy of maps $F' \to G$ is zero. Thus $RHom(F,G) = Hom(F',G)$ is 1-dimensional whereas $Hom(F,G) = 0$, so again the former is not a quotient of the latter. $\endgroup$
    – Tim Campion
    Oct 13, 2020 at 2:52
  • 4
    $\begingroup$ Prop. 2.15 and 2.20 from this paper of mine numdam.org/item/BSMF_2010__138_3_317_0 tell us that the functor $Ho(Fun(T,M))\to Fun(T,Ho(M))$ full, essentially surjective and conservative for any model category $M$ and any tree $T$. Hence it induces a bijection on the collections of isomorphism classes. $\endgroup$ Oct 13, 2020 at 5:51

Let $T$ be a well-founded poset and $k$ a field. Let $H: Ch_k \to Ho(Ch_k) = Gr_k$ be the homology functor and $\iota: Gr_k \to Ch_k$ be the canonical section which sends a graded vector space to the corresponding chain complex with zero differential.

Proposition 1: For any $F: T \to Ch_k$, $F$ and $\iota H(F)$ are quasi-isomorphic.

In particular, if $H(F) \cong H(G)$, then $F \simeq \iota H(F) \cong \iota H(G) \simeq G$.

Proof: In the Reedy = injective model structure on $Fun(T,Ch_k)$, fibrations and weak equivalences are levelwise, and an object is cofibrant (and thus bifibrant) if and only if each morphism of $T$ is carried to a monomorphism. We may assume that $F \in Fun(T,Ch_k)$ is cofibrant and prove the more specific claim that there is a quasi-isomorphism $F \to \iota H(F)$. By induction on the structure of $T$, we are reduced to the following:

Lemma 2: Let $X \to Y$ be a monomorphism in $Ch_k$. If $X \to \iota H(X)$ is a quasi-isomorphism, then there exists a quasi-isomorphism $Y \to \iota H(Y)$ forming a commutative square.

Proof: We may decompose $X \to Y$ into a series of cell attachments, and construct the map $Y \to \iota H(Y)$ by induction, one cell at a time. There are two cases. In the first case we have $Y = X \oplus k[n]$, in which case we have $H(Y) = H(X) \oplus k[n]$, and we extend using the identity on $k[n]$. In the second case we have $Y = X \cup_z w$ where $z$ is a cycle representing a nonzero homology class of $X$ and $w$ is a cell we are attaching to kill it. In this case we have $H(Y) = H(X)/v$, and we can extend via the zero map on $w$.

Added in response to comment: It's worth noting that although the the proposition only applies directly to tree-shaped diagrams, we can leverage it to understand certain more complicated diagrams. For example:

Let $\mathcal C$ be the "marked category" $$\require{AMScd} \begin{CD} a @>>> b\\ @VVV @VVV\\ 0 @>>> c \end{CD}$$

with the understanding that $Fun(\mathcal C,Ch_k)$ means functors $\mathcal C \to Ch_k$ carrying $0$ to an acyclic chain complex. Then $Fun(\mathcal C,Ch_k)$ with levelwise weak equivalences models "maps $A \xrightarrow f B \xrightarrow g C$ equipped with a nullhomotopy of $gf$".

Lemma 3: If $\mathcal D$ is/presents a stable $\infty$-category, then the $\infty$-category (presented by) $Fun(\mathcal C, \mathcal D)$ is equivalent to the $\infty$-category (presented by) $Fun(T,\mathcal D)$ where $T$ is the tree $D \leftarrow F \to B$.

Proof: Consider the following diagram, where the rows and columns are fiber sequences:

$$\require{AMScd} \begin{CD} A @>>> F @>>> D \\ @V{=}VV @VVV @VVV \\ A @>>> B @>>> Q \\ @VVV @VVV @VVV \\ 0 @>>> C @>{=}>> C \end{CD}$$

The diagram can be uniquely reconstructed from either the bottom left square, or the top right square subject to the requirement that it be an exact square, or from the part $D \leftarrow F \to B$ subject to no requirements.

Proposition 4: In $Fun(\mathcal C, Ch_k)$, a diagram is quasi-isomorphic to its homology. Consequently, if the homologies of two such diagrams are isomorphic, then the diagrams are quasi-isomorphic.

Proof: By Proposition 1, this is true for $Fun(T,Ch_k)$. Start with a model of the top right square such that $F \to D$, $F \to B$, and $B \oplus D \to Q$ are monomorphisms. By Proposition 1, we have a commuting square of quasi-isomorphisms from $F \to D \oplus B$ to the corresponding map of homologies. By Lemma 2, we may extend this to a quasi-isomorphism from $Q$ to its homology, in a natural way. We think of this as a natural transformation from the top-right square of the above grid to the corresponding grid of homologies, and we seek to fill out the remaining components of a natural transformation between grids.

Since $D \to Q$ is a monomorphism, we may use Lemma 2 again to extend this natural family of quasi-isomorphisms to $C$, and hence to the lower-right square. Modeling $A$ by the dual mapping cone of $F \to D$, we obtain a model of $A \to F$ which is epic. By the dual of Lemma 2, we may extend our natural family of quasi-isomorphisms to $A$ and hence to the upper-left square. To model the lower-right square, take an arbitrary factorization through an acyclic complex, which comes with a unique quasi-isomorphism to its homology. The remaining naturality squares all necessarily commute, and the naturality of the quasi-isomorphisms out of the lower-left square is precisely what we are looking for.

  • $\begingroup$ Did you mean Fun(T,Ch_k) as opposed to Fun(Ch_k,T)? $\endgroup$ Oct 12, 2020 at 19:33
  • $\begingroup$ @DmitriPavlov Fixed, thanks! $\endgroup$
    – Tim Campion
    Oct 12, 2020 at 19:34
  • $\begingroup$ @DmitriPavlov: Thank you so much, Dmitri. Do you think that the classification of the diagram of the form $A\langle n \rangle_\ast \stackrel{f}{\to} B\langle n \rangle_\ast \stackrel{g}{\to} C \langle n \rangle_\ast$ with $g\circ f \sim 0$, can be recovered form the $ho(Ch_\mathbb{Q})$ and the classification of the diagram of the type $E\langle n \rangle_\ast \to 0 \to \Sigma E\langle n \rangle_\ast$? Here $A\langle n \rangle_\ast$ is the chain complex concentrated in degree $n$ and $A\langle n \rangle_n=A.$ $\endgroup$ Oct 12, 2020 at 20:00
  • 2
    $\begingroup$ @SurojitGhosh Not sure if you were talking to me, and not sure I understood the question correctly, but I added an edit inspired by your comment. $\endgroup$
    – Tim Campion
    Oct 12, 2020 at 21:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.