Show that every homeomorphism is a local homeomorphism

This is the definition of local homeomorphism I was given:

" Let $$(X,\tau)$$ and $$(Y,\tau_1)$$ be topological spaces. A map $$f:X \to Y$$ is said to be a local homeomorphism if each point $$x \in X$$ has an open neighbourhood $$U$$ such that the restriction of $$f$$ to $$U$$ maps $$U$$ homeomorphically onto an open subspace $$V$$ of $$(Y,\tau_1)$$; that is, if the topology induced on $$U$$ by $$\tau$$ is $$\tau_2$$ and the topology induced on $$V=f(U)$$ by $$\tau_1$$ is $$\tau_3$$, then $$f$$ is a homeomorphism of $$(U,\tau_2)$$ onto $$(V,\tau_3)$$. The topological space $$(X,\tau)$$ is said to be locally homeomorphic to $$(Y,\tau_1)$$ if there exists a local homeomorphism of $$(X,\tau)$$ into $$(Y,\tau_1)$$."

Now to show that every homeomorphism is a local homeomorphism, the main idea is that each open neighborhood is an open set and the image is also an open set. Then the induced subspace of both are actually just open sets from the original topologies and we take the same function from the original homeomorphism.

• Yes. It’s not just the main idea: for all practical purposes it’s the proof. (In fact you can simply take $X$ itself to be the open nbhd of $x$.) Jan 18, 2021 at 3:40

What you wrote out is a proof but another (and in my opinion simpler proof) is for each point to take the entire space, $$X$$, as your open neighborhood. The restriction of $$f$$ to $$X$$ is just $$f$$ which is a homeomorphism by assumption.