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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.

UCL
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2 Answers2

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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 $H(t,s)=\lambda(t)(1-s)$.

Frederik
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I think this is a counterexample: take $X=S^1$ and $$ F(t, r) = \big(\sin(32\pi r(r-1)t(t-1)), \cos(32\pi r(r-1)t(t -1))\big) $$ This is an homotopy between the constant path (at r=0) and the constant path (at r=1) but the image is the circle which is not simply connected, right?

Quimey
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