TITLE: ¿Is the image of a path homotopy simply connected?
QUESTION [6 upvotes]: I was just introduced to algebraic topology in class, we've learned about homotopies, covering maps and deformation retracts. We've also calculated some fundamental groups.
I was wondering, asume you have a loop $\alpha : I \longrightarrow X$ based on $x_0 \in X$ which is path homotopic to the trivial loop $\varepsilon_{x_0}$ via a homotopy
$F:I \times I \longrightarrow X$. Can we deduce that $\text{Im}(F)$ is simply connected?
From now I assume $\text{Im}(F)=X$
My intuition tells me that it should be, we certainly know that $X$ is path connected. And the fact that $F$ is defined on $X$ means that you can "sweep a rope" across $X$ and it doesn't get stuck on anything. I thought $X$ might even be contractible but i had no luck formalizing either of those intuitions.
Does anybody know any result that would prove this, even if I can't really grasp it know? Or maybe give a counterexample. But I can't think of any.
Thanks in advance.

REPLY [4 votes]: This is not the case, and there are trivial counterexamples. For example there exists surjective loops in $S^1$ which are path homotopic to the constant path. An examples is $\lambda:I/\partial I \to I/\partial I(\cong S^1)$ given by
$$
\lambda(t)=
\begin{cases}
2t & 0\leq t \leq 1/2 \\
2(1-t) & 1/2\leq t \leq 1
\end{cases}
$$
A path null-homotopy $H:I/\partial I\times I\to I/\partial I$ is given by $H(t,s)=\lambda(t)(1-s)$.