Acyclic model type result: existence of a chain homotopy If $\sigma: \Delta_n \to X$, define $\overline{\sigma}: \Delta_n \to X$ by$$\overline{\sigma}(t_0, \dots, t_n) := \sigma(t_n, \dots, t_0).$$Define a map $T: C_n(X) \to C_n(X)$ by $T(\sigma) := (-1)^{n(n+1)/2}\overline{\sigma}$.


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*Show that $T$ is a chain map.

*Show (without constructing it explicitly) that there exists a chain homotopy from $T$ to the identity.


Progress so far. For the first part, the proof that $\partial T = T \partial$ is a straightforward matter of comparing signs. But I am not sure on how to approach the second part.
 A: A reference for acyclic models theorems:

Samuel Eilenberg and Saunders MacLane. Acyclic models. Amer. J. Math., 1953, 75, 189-199

There's also Wikipedia if you don't want to read their proof. I will rephrase their results in a way that's applicable here.
Suppose that $\mathsf{C}$ is a category, and $\mathsf{M} \subset \mathsf{C}$ is a set of objects, called model objects. If $T : \mathsf{C} \to \mathsf{Ab}$ is a functor from $\mathsf{C}$ to the category of abelian groups, it is said to be representable if, roughly, for all objects $A \in \mathsf{C}$, the group $T(A)$ is "free" on the $T(M)$ for $M \in \mathsf{M}$. It means that every $T(A)$ has a (natural in $A$) basis of elements of the form $T(\varphi)(m)$ where $M \in \mathsf{M}$, $m \in M$ and $\varphi : M \to A$.
Then the Theorem II of the paper I cited implies the following fact. Suppose that if $K, L : \mathsf{C} \to \mathsf{Ch}$ are functors from $\mathsf{C}$ to the category of chain complexes, and suppose that $f_0 : K_0 \to L_0$ is a natural transformation. Then if $K_n$ is representable for all $n \ge 1$ and $H_n(L(M)) = 0$ for all $n \ge 1$, $M \in \mathsf{M}$, then there exists a unique up to homotopy extension $f : K \to L$ of $f_0$.

Now how does this all apply here? The category $\mathsf{C} = \mathsf{Top}$ is the category of topological spaces. The set of models objects is the set of standard simplexes: $\mathsf{M} = \{ \Delta^n \mid n \ge 0 \}$. The two functors $K = L : \mathsf{Top} \to \mathsf{Ch}$ will actually be equal to the singular chains functor $C$.
The fact that $H_n(C(M)) = H_n(M; \mathbb{Z}) = 0$ for $n \ge 1$ and $M \in \mathsf{M}$ is a simple consequence of the fact that the standard simplexes are contractible, a standard fact in algebraic topology.
The fact that the functor $C_n$ is representable means that for every space $X$, every chain $x \in C_n(X)$ can be written as a linear combination of $C_n(\sigma)(c)$ for some $c \in C_n(\Delta^m)$ and $\sigma : \Delta^m \to X$. This is basically by the definition of the singular chains: take $i_n \in C_n(\Delta^n)$ to be the identity, then $C_n(\sigma)(i_n)$ is equal to $\sigma$ seen as an element of $C_n(X)$. And $C_n(X)$ has a basis consisting of such $\sigma : \Delta^n \to X$.
Now you have two natural transformations $C \to C$: the identity $\operatorname{id}$, and $T$ (check that $T$ is natural). Let $f_0 = T = \operatorname{id} : C_0 \to C_0$ (by definition of $T$, it agrees with the identity in dimension zero). But the theorem states that $f_0$ has a unique up to homotopy extension $C \to C$; since $T$ and $\operatorname{id}$ are both such extensions, there must exist a homotopy between the two.
