Let $M$ be a non-compact connected surface. One can assume that $\partial M = \varnothing$. It follows from the results of the [paper](https://projecteuclid.org/euclid.acta/1485889458) - Epstein, D. B. A. *Curves on 2-manifolds and isotopies*. Acta Math. 115 1966 83–107. that 1) $\pi_1 M$ is **locally free**, i.e. every finitely generated subgroup $G$ of $\pi_1 M$ is free; 2) moreover, $\pi_1 M$ is a union of an increasing countable sequence of finitely generated free subgroups. However, I do not know how to deduce that $\pi_1 M$ is free. That paper contains elementary proofs of basic results on curves on surfaces and homemorphisms surfaces. Recall that a connected subsurface $N \subset M$ is **incompressible** if the homomorphism $\pi_1 N \to \pi_1 M$ is injective. **Lemma 2.2 of Epstein's paper.** *Let $X \subset M$ be a compact subset and $G$ be a finitely generated subgroup of $\pi_1 M$. Then there is a compact incompressible subsurface $N \subset M$ such that* - $X \subset int(N)$ - $G \subset \pi_1 N \subset \pi_1 M$. **Corollary 1**. *$\pi_1 M$ is locally free. *Proof*. Indeed, since $N$ is compact and has non-empty boundary, $N$ can be deformed onto a finite graph, and therefore $\pi_1 N$ is free and contains $G$. Hence, by [Nielsen–Schreier theorem](https://en.wikipedia.org/wiki/Nielsen%E2%80%93Schreier_theorem), $G$ is free as well. **Corollary 2**. $\pi_1 M$ is a union of an increasing countable sequence of finitely generated free subgroups. *Proof*. Represent $M$ as a countable union of compact subsets $X_1 \subset X_2 \subset \cdots $ such that $M = \cup_i X_i$. Let also $G_0 = 1$ be the unit subgroup of $\pi_1 M$, and $N_0 \subset M$ be an incompressible subsurface such that - $X_0 \subset int(N_0)$ and $G_0 \subset \pi_1 N_0 \subset \pi_1 M$. Denote $G_1 = \pi_1 N_0$, and let $N_1 \subset M$ be an incompressible subsurface such that - $X_1 \subset int(N_1)$ and $G_1 \subset \pi_1 N_1 \subset \pi_1 M$, repeating this process we will obtain that an increasing sequence of incompressible compact subsurfaces $N_0 \subset N_1 \cdots $ such that $M = \cup_i N_i$. Since every loop in $M$ is contained in some compact subset and therefore in some $N_i$, it follows that $\pi_1 M $ is a union of its finitely generated free subgroups $\pi_1 N_0 \subset \pi_1 N_1 \subset \cdots $