I am reading a paper on closed curves of a closed-oriented surface, where the following Fact comes without proof.
Fact: Let $\Sigma$ be closed connected orientable surface. Let $\widehat\pi(\Sigma)$ be the free homotopy classes of loops on $\Sigma$. Consider $x,y\in \widehat\pi(\Sigma)$. Then there is a representative $\alpha$ of $x$ and a representative $\beta$ of $y$ such that the following hold:
$(1)$ Both $\alpha$ and $\beta$ are the images of some smooth maps $\Bbb S^1\to \Sigma$.
$(2)$ Each point of $\alpha\cap \beta$ is a point of transversal intersection, i.e., each point $p\in\alpha\cap \beta$ has a small open neighborhood $U_p$ such that $U_p\cap (\alpha\cup \beta)$ is homeomorphic to $(\Bbb R\times 0)\cup (0\times \Bbb R)$.
My question is how to prove the above Fact. My idea is to use geodesic representatives of $x$ and $y$.
It's probably not proved in that paper because it is a standard elementary theorem of differential topology known as the transversality theorem.
You should be able to find the proof in most differential topology textbooks. I also found these lecture notes by a search for transversality theorem.