homeomorphic image of an simply connected set

Question

How can I prove, that the homeomorphic image of an simply connected set is also simply connected?

It is Convex $\to$ radially $\to$ simply connected.

I have no idea how to prove that.

Answer 1

Homeomorphisms preserve all topological aspects of a space, essentially by definition. Concretely, assume $h:X \to Y$ is a homeomorphism and $X$ is simply connected. This means that $X$ is path-connected and that all the loops in $X$ are contractible.

To see that $Y$ is path-connected, pick any two points $y_1,y_2 \in Y$. Then since $X$ is path-connected, there exists a path $p:I \to X$ (where $I = [0,1]$) starting at $h^{-1}(y_1)$ and ending at $h^{-1}(y_2)$. Then $h \circ p:I \to Y$ is a path starting at $y_1$ and ending at $y_2$.

To see that any loop in $Y$ is contractible, pick a loop $l:S^1 \to Y$ based at a point $y_0 \in Y$. Then $h^{-1}\circ l:S^1\to X$ is a loop in $X$ based at $x_0 = h^{-1}(y_0) \in X$, which is contractible by hypothesis. This means that there exists a homotopy $H:S^1 \times I \to X$ such that $H(t, 0) = h^{-1}(l(t))$ and $H(t,1) = x_0$. Then $h \circ H:S^1 \times I \to Y$ provides the necessary homotopy that contracts $l$ to $y_0$.