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This is sort of a lemma I'm trying to prove for a larger proof. It seems intuitively true: if a space has trivial fundamental group, any two loops based at a point are homotopic. A subspace of such a space should have the same property, since loops based at a point must lie in the same path component as the point. That path component is path connected, and any two loops in it are homotopic, so the path component is simply connected as well. I'm having trouble making the "path component has trivial fundamental group" part rigorous though. Can anyone give me a nudge in the right direction? My idea so far is to use the existence of paths connecting points on any two loops to make a homotopy between them but I'm not sure how to make that formal.

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    $\begingroup$ Take any simply connected subset of $\mathbb{R}^2$ with non-empty interior and remove a point in the interior. Or take the unit ball in $\mathbb{R}^3$ and remove the $x$-axis. $\endgroup$ – copper.hat Sep 10 '13 at 17:49
  • $\begingroup$ Take any space $X$ and form the cone $CX$, which is contractible, and X is a subspace. $\endgroup$ – Justin Young Sep 11 '13 at 8:06
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Take the unit sphere $S^1$ as a subspace of the simply connected space $\Bbb R^2.$

Note that $S^1$ is not a retract of $\Bbb R^2$. Otherwise, it had to be simply connected, too.

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The question from the title was already answered, I will answer the question form the "body" of the post: we will show that path components of a space with trivial fundamental group have trivial fundamental groups.

Consider any path component $C$ of the space $X$, $x_0\in C$ and any loop $\omega$ based at $x_0$. Because the fundamental group is trivial, we have a homotopy $h:S^1\times [0,1]\to X$ between $\omega$ and the constant loop. Because $S^1\times [0,1]$ is path connected and images of path connected spaces are path connected, we conclude that $h(S^1\times[0,1])$ must lie in $C$ (do not intersect any other component of the space). Thus, the result follows.

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