Any element $\alpha \in \pi_1(M)$ can be represented by a smooth closed path without self-intersections

Question

Like the title says, any element $\alpha \in \pi_1(M)$ can be represented by a smooth closed path without self-intersections, if the manifold $M$ is of dimension $3$ or higher, as "there is enough room to get rid of the self-intersections in any closed path by means of arbitrarily small deformations". This argument makes sense, but I don't know how I would properly write that down in mathematical terms.

Answer 1

Here is a sketch. I can add more details, but this would require some basic understanding of transversality on your part.

Let $M, N$ be smooth manifolds of dimensions $m, n$ respectively, so that $n> 2m$ and $M$ is compact. Then given any smooth map $f: M\to N$, we obtain the map of squares: $$ f^2:=f\times f: M\times M\to N\times N. $$
Now, one can make a small perturbation $g$ of $f$ to make $g^2$ transversal to the diagonal $\Delta\subset N\times N$. The dimension count yields $2m=dim(M\times M)< dim(\Delta)$. Hence, the perturbed map satisfies the property that $im(g^2)\cap \Delta=\emptyset$. Therefore, $g$ is 1-1. A similar argument involving $df$ ensures existence of a small perturbation which is also an immersion, hence, an embedding. Furthermore, for a sufficiently small perturbation, $g$ is homotopic to $f$. Now, apply this to $M=S^1$.