# Proof of Theorem 6.8 in the paper “Singular homology of abstract algebraic varieties”

Background: Theorem 6.8 in Suslin and Voevodsky's article "Singular homology of abstract algebraic varieties" states that there is an isomorphism between effective relative zero cycles $z_0^c(Z)^{eff}(S)$ and the monoid $Hom(S,\coprod_{d=0}^{\infty} Sym^d(Z))[1/p]$, where $Z$ is any algebraic scheme and $S$ is a normal connected algebraic scheme.

Constructing the isomorphisms: The map $z_0^c(Z)^eff(S) \to Hom(S,\coprod_{d=0}^{\infty} Sym^d(Z))[1/p]$ is given as follows: Suppose that $X \to Z \times S$ is a closed embedding such that the composition $X \to Z \times S \to S$ is finite and surjective. Then it is proved in the paper that we have a composition of morphisms $$f_X: = (S \to Sym^d(X/S) \to Sym^d(Z) \times S \to Sym^d(Z))$$ where the first morphism $S \to Sym^d(X/S)$ is a section of the projection $Sym^d(X/S) \to S$. Now the map $$z_0^c(Z)^{eff}(S) \to Hom(S,\coprod_{d=0}^{\infty} Sym^d(Z))[1/p]$$ is given by $$X \mapsto f_X.$$ The inverse map is given as follows: Consider the presheaf $$T \mapsto \mathbb{Z}[1/p](Hom_{Sch/k}(T,Z))$$ and let $\mathbb{Z}[1/p]_{qfh}(Z)$ denote the sheafification with respect to the qfh-topology. It turns out that the sheaf $\mathbb{Z}[1/p]_{qfh}(Z)$ is isomorphic to the sheaf of relative zero cycles $z_0^c(Z)$. Letting $q: Z^d \to Sym^d(Z)$ denote the projection, there is is a unique element $$u_d \in \mathbb{Z}[1/p]_{qfh}(Z)(Sym^d(Z))$$ such that $q^{*}(u_d) = \sum_i \pi_i \in \mathbb{Z}[1/p](Z)(Z^d)$ where $\pi_i: Z^d \to Z$ denotes the projection onto the i'th factor. Now given any $$f: S \to Sym^d(Z)$$ we map this to the element $$f^{*}(u_d) \in \mathbb{Z}[1/p]_{qfh}(Z)(S) = z_0^c(Z)(S)$$. The article now states that one verifies easily that $f^{*}(u_d)$ is an effective relative zero cycle and that the resulting map $$Hom(S,\coprod_{d=0}^{\infty}Sym^d(Z))[1/p] \to z_0^c(Z)^{eff}(S)$$ is an inverse to the one constructed before.

My question: How does one verify that these two maps constructed above are infact inverses to each other?

What I have attempted: I have tried showing that if $X \to Z \times S$ is a relative zero cycle then $f_X^{*}(u_d) = X$. Which using that $S \to Sym^d(X/S)$ is a section should be equivalent to showing that the class of $$\sum_i \pi_i|_{X^d/S \times_S Y} \in \mathbb{Z}[1/p]_{qfh}(Z)(X^d/S \times_S Y)$$ coincides with the class of $$\sum_{r \in Hom_S(Y,X)} p_1|_{X} \circ r|_{X^d/S \times_S Y} \in \mathbb{Z}[1/p]_{qfh}(Z)(X^d/S \times_S Y)$$ where $Y \to S$ is a normalization of $S$ in a field extension of $k(S)$ containing $k(X)$ and $p_1|_{X}$ is the composition $X \to Z \times S \to Z$. I cannot see why this should be the case though, and I have no idea how to prove that the opposite composition yields identity.

The proof that the maps compose to the identity is rather long, so I will just say a few words of how this is done here: First of all the element $u_d \in \mathbb{Z}[1/p]_{qfh}(Z)(Sym^d(Z))$ can be described explicitly in terms of the covering $\{Z^d \to Sym^d(Z)\}$ and the class of the element $\sum_{i=1}^d \pi_i$. We then compute what this pulls back to in $\mathbb{Z}[1/p]_{qfh}(Z)(S)$ and then we pull this back to $Y$. It is enough to show that what we now have coincides with the image of $X$ in $\mathbb{Z}[1/p]_{qfh}(Z)(Y)$. Producing the required $qfh$-covering is done by using that a morphism with the properties of Lemma 5.1 in the article is preserved by base change and we will also need some additional properties of the section of symmetric powers which are not mentioned in the article, but the key needed result resembles proposition 4.7 in the book "Linear determinants with applications to the picard scheme of a family of algebraic curves" by Iversen.
After we have showed that $f_{X}^{*}(u_d) = X$, we only need to show that if $f \ne g$ then $f^{*}(u_d) \ne g^{*}(u_d)$ which more or less just follows from universal properties.