Trees in chain complexes

Question

$\DeclareMathOperator{\Ch}{\mathit{Ch}}$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.

Answer 1

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.

Answer 2

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 $w$.

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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.