$\DeclareMathOperator{\Ch}{\mathit{Ch}}$Let $\Ch_\mathbb{Q}$ denotes 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.

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

Let $Ch_\mathbb{Q}$ denotes 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 $ho(Ch_{\mathbb{Q}})$?

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 $dim(ker(H_n(f_\ast)))_n$.

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

$\DeclareMathOperator{\Ch}{\mathit{Ch}}$Let $\Ch_\mathbb{Q}$ denotes 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.