Proving statements about homeomorphisms are equivalent 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.
 A: If you want to show that $\phi_1, \ldots, \phi_n$ are equivalent, you need to prove enough of the statements $\eta_{ij} \equiv \phi_i \Rightarrow \phi_j$, so that the $\eta_{ij}$ viewed as edges on a directed graph with vertices $1, \ldots, n$ gives a (strongly) connected graph. In your example, you have produced a cyclic graph that contains all the vertices, so that is fine. As for finding the most efficient route, that is a value judgment that depends on the particular problem. In your example, the cyclic graph looks quite efficient to me, as it minimises the number of edges in the graph. However, that may not be the case in other examples. Sometimes (but not in this example) it is advantageous to introduce more $\phi_i$ to make the implications $\eta_{ij}$ easier to prove.
A: The open map and closed map conditions are very similar to each other, so it makes sense to prove both at the same time give $(a)$ as a hypothesis. When proving $a$ given either $(b)$ or $(c)$ it makes sense to pick the one that is friendliest given the definition of continuity that you are using. I'm using the inverse images of open sets are open definition of continuity here.
As Rob Arthan points out, this isn't quite enough, we also need to show that $(b)$ implies $(c)$ in order to get the missing $(b)$ implies $(a)$ edge. I include an argument for this case initially and so my answer was incorrect.
As an addendum, this kind of error is precisely why it's a good idea to decide on the structure of implications up front and keep it simple: $A \implies B \implies C \implies A$ and $A \iff B \iff C$ are both good choices.

I think the easiest way to prove this is to show that $(a) \implies (b)$, $(a) \implies (c)$ and $(c) \implies (a)$.
First a word on notation, if $f$ is a function, let $f^\to$ be the function that sends $X' \subset X$ to the direct image of $X$ under $f$. Likewise, let $f^\leftarrow$ be the function that sends $Y' \subset Y$ to its inverse image.
I'll say a function $f$ is continuous if and only if $f^{\leftarrow}$ is an open map. It is a theorem that $f$ is continuous, then $f^{\rightarrow}$ is a closed map.

By hypothesis, $f$ is continuous. Therefore $f^{\leftarrow}$ is an open map and $f^{\leftarrow}$ is a closed map.
Let us further suppose that $f$ is a homeomorphism. That means that it has a continuous inverse. Let $g$ be the inverse of $f$.
$g^{\leftarrow}$ is an open map and a closed map by definition of continuity and our theorem about continuity.
However, since $g$ is the inverse of $f$, $g^{\leftarrow}$ is equal to $f^\to$. Therefore $f^\to$ is an open map and a closed map. $f$ has an inverse, therefore it's a bijection.
This gives us $(b)$ and $(c)$ assuming $(a)$ as a hypothesis.
Suppose that $f$ is an open bijection. This means that $f^{\to}$ is an open map and $f$ is bijective. Thus the inverse of $f$ exists, call it $g$.
$g$ is continuous with a continuous inverse, since $f$ is constrained to be continuous by hypothesis.
Therefore $g$ is a homeomorphism.
Therefore $f$ is a homeomorphism.

Suppose $f$ is a closed bijection. This means that $f$ is a bijection and $f^{\to}$ is a closed map.
Since $f$ is a bijection, it has an inverse $g$.
Since $g^{\rightarrow}$ is a closed map, $g$ is continuous.
Since $g$ is continuous, $g^{\rightarrow}$ is an open map.
Therefore $f^{\to}$ is an open map.
Therefore $f$ is an open bijection.
A: Proving a ring of implications is often the most efficient way to proceed, and it works fine here, but I’d probably proceed a bit differently with this result. In all three conditions the function in question is a bijection, so the theorem boils down to showing that if $f$ is a bijection, the following are equivalent: $f^{-1}$ is continuous, $f$ is closed, and $f$ is open.
Since $f$ is a bijection, then $f[X\setminus A]=Y\setminus f[A]$ for any $A\subseteq X$, and it is an immediate consequence of this that $f$ is closed if and only if it is open. (You may well want to fill in the details, but they really are very easy.) Moreover, $f^{-1}$ is continuous if and only if $f$ is open. The first observation yields the equivalence of (b) and (c), and the second yields the equivalence of (a) and (c). Since all three conditions are equivalent to (c), they are mutually equivalent.
