Above proof, [Topology, J.Munkres (Part 2 Algebraic topology)]
I cannot show that the map $\pi: S^1\times I\to B^2$ given by $\pi(x,t)=(1-t)x$ is continuous, closed and surjective, but is not open.
Any help?
2026-02-23 12:09:21.1771848561
continuous, closed and surjective not open.
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Consider $\{(x,y)\in S^1\,|\,y>0\}\times I$. It is an open subset of $S^1\times I$, but its image under $\pi$ is$$\{(x,y)\in B^2\,|\,y>0\}\cup\{(0,0)\},$$which is not an open subset of $B^2$. Therefore $\pi$ is not an open map. But it is closed, since it is continuous and its domain is compact.