Is an arc $f:[0,1]\to X $ with base point $x_0\in X$ a surjective function?

Question

Let be X a topological space. Is an arc $f:[0,1]\to X $ with base point $x_0\in X$ a surjective function? I know that it is a continuous function. Can you help me?

Answer 1

No, not at all. You can just define the trivial loop by $f(x) = x_0$. If your space $X$ has more than $2$ elements this will not be surjective. In general you will pretty much never have surjective loops. To give an interesting looking example, look at the loop that I have drawn on this Möbius strip:

Clearly this does not cover the entire space.

P.S.: For some reason if I rescale the picture to medium size, the white background gets inverted. Sorry if the image is a bit huge now.

Answer 2

No, it's not surjective. Take, $X=\Bbb R$ with usual topology and $f:[0,1]\to X$ by $f(x)=x$ for all $x\in [0,1]$ the clearly $f$ is continuous and an arc too. But the point $2\in X$ has no preimage in $[0,1]$.