# Simply connected subset of $\mathbb R^3$

Let $C$ be the closed unit cube in $\mathbb R^3$, and let $A$ be one face of the cube $C$ (say the face above and parallel to $xy$-plane). Let $U\subset\mathbb R^2$ be open and path-connected such that $U\subseteq A$.

I claim that the space $X:=(C-A)\cup U$ is simply connected, by dragging any loop in $X$ into $int( C)$ which is convex subset of $\mathbb R^3$ i.e contactable.

My Question is: How to prove that formally? Can you find a retraction (homotopy map) which shows that $X$ is homotopy equivalent to $C-A$ (or equivalently that $X$ is contactable)?

• Can't you just uniformly scale the path? – user7530 Oct 19 '13 at 14:09
• @user7530 What do you mean by scaling paths? – Ronald Oct 19 '13 at 14:11
• Center the cube at the origin. If $\gamma(t)$ is a loop on $X$, isn't $(1-s)\gamma(t)$ in $X$ for $0\leq s\leq 1$, and therefore a contraction of $\gamma$ to the origin? – user7530 Oct 19 '13 at 14:13
• $X$ is star-shaped with respect to the origin. – Daniel Fischer Oct 19 '13 at 14:17
• @Ronald $A$ is path-connected by definition. I suppose you meant $U$ ; if so it does not matter : any path in $U$ is a path in one of the path-connected component of $U$, so $U$ can be taken path-connected without loss of generality. – Pece Oct 19 '13 at 15:24

$X$ is starshaped with the center of the star at any interior point of $C$, and so $X$ is contractible.