# A proof of the Dold$-$Thom theorem via factorization homology

## TL;DR

This paper offers a new, direct proof of the Dold–Thom theorem by identifying the infinite symmetric product with an instance of factorization homology, avoiding traditional quasi-fibration arguments.

## Contribution

It introduces a novel proof technique for the Dold–Thom theorem using factorization homology, providing a more straightforward approach.

## Key findings

- Identifies the infinite symmetric product as an instance of factorization homology.
- Provides a direct proof of the Dold–Thom theorem without quasi-fibration arguments.
- Enhances understanding of the relationship between symmetric products and factorization homology.

## Abstract

The Dold$-$Thom theorem states that for a sufficiently nice topological space, M, there is an isomorphism between the homotopy groups of the infinite symmetric product of M and the homology groups of M itself. The crux of most known proofs of this is to check that a certain map is a quasi-fibration. It is our goal to present a more direct proof of the Dold$-$Thom theorem which does appeal to any such fact. The heart of our proof lies in identification of the infinite symmetric product as an instance of factorization homology.