I am trying to prove the following equivalences. What is a good way to prove statements in a chain of statements like this are equivalent. By both taking the easiest, most efficient route for proving equivalency, and also making sure each equivalence has indeed been proved? Is there any systematic way for doing this?
If $f:X \rightarrow Y$ is continuous, the following are equivalent:$(a)$$f$ is a homeomorphism$(b)$$f$ is a closed bijection.$(c)$$f$ is an open bijection.
Attempt:I am struggling to find the correct chain of equivalences that would make this the easiest.
I will try $(a)\implies (b) \implies (c) \implies (a)$ Would I also need to show $(c) \implies (b)$ and $(b) \implies (a)$. Aside from the proof, what is a general strategy for determining a sequence of equivalences, that will show each statement implies the other?
$(a) \implies (b)$If $f$ is a homeomorphism, $f$ is certainly bijective and for any closed set $A$ in $X$, $(f^{-1})^{-1}(A)=f(A)$ so that $f$ is a closed map since $f^{-1}$ is continuous.
$(b) \implies (c)$If $f$ is a closed bijection, for any closed set $A$, let $U=X-A$. Then $f(U)=(f^{-1})^{-1}(X-A)=(f^{-1})^{-1}(X)-(f^{-1})^{-1}(A)=Y-f(A)$ and so $f(U)$ is open.
$(c) \implies (a)$If $f$ is an open bijection for any open set $U$ in $X$, $f(U)$ is open in $Y$. So $f^{-1}$ is continuous. So $f$ is a homeomorphism.
