what is a path that cover all of $S^n$?

Question

Here is the meaning of "cover" which I can't understand:
Prove that if $n\ge 2$, then $S^n$ is simply connected. hint: Use Exercise 2.5 to show that every loop in $S^n$ is homotopic to a loop that does not cover all of $S^n$.
Here is Exercise 2.5 which I can use it:
Show that any path in $S^n$ is homotopic with endpoints fixed to a polygonal path on $S^n$, where "polygonal" is now interpreted to mean that the path is formed from arcs lying on great circles of $S^n$.

Answer 1

In this context for a path or loop in a topological space $X$ to "cover" $X$ simply means that the continuous function $f : [0,1] \to X$ or $f : S^1 \to X$ which defines the path or loop is surjective: for each $y \in X$ there exists $x \in \text{domain}(f)$ such that $f(x)=y$. This is quite different from other topological uses of the word "cover".

If $X$ is a manifold of dimension $\ge 2$ and if $f$ is smooth, or piecewise smooth (e.g. if $X$ is contained in $\mathbb{R}^m$ and $f$ is polygonal) then $f$ is not surjective. But even for certain higher dimensional compact spaces like $X=S^n$ there do exist surjective continuous paths and loops, as is shown in the link provided by @PeterFranek.