Equivalent definitions of "Homeomorphisms" I found that the equivalent definition of a homeomorphism is as follows:
Let X and Y be two topological spaces. A map $f : X \rightarrow Y$ is said to be homeomorphism if $f$ is bijective and $f$ and $f^{-1}$ is continuous.
Now I found a equivalent definition of a homeomorphism which is as follows: $U$ is open in $X$ iff $f(U)$ is open in $Y$.
I got stuck to prove the above.
$\textbf{My Attempt:}$
Consider $f : X \rightarrow Y$ is continuous. Since $f(U)$ is is open in $Y$, by the definition of continuity $f^{-1}(f(U))$ is open in X. Now since $f, f^{-1}$ are bijection, then $f^{-1}(f(U)) = U$ (in general,  $U$ is a subset of $f^{-1}(f(U))$, and equality holds if $f$ is a injective map (bijective, in particular)). Thus, $U$ is open in $X$.
Now we will have to prove $U$ is open in $X$, then $f(U)$ is open in $Y$. I have an idea that for this we should take the map $f^{-1}: Y \rightarrow X$. Now if $U$ is open in $X$, then by the definition of continuity we have $(f^{-1}) ^{-1} (U)$ is open in $Y$.
Does $(f^{-1}) ^{-1} (U)$ mean $f(U)$?
Now  I have now idea. Please help me.
 A: As Kavi Rama Murthy already said in the comments, yes $(f^{-1})^{-1}(U)=f(U)$. I will give you a simple hint on how to prove it. By definition:
$$(f^{-1})^{-1}(U)=\{y\in Y : f^{-1}(y)\in U\}$$
And finally $f^{-1}(y)\in U \iff y=f(x)$ for $x\in U$ (again, by definition). I think you can prove the rest.
However, note that your proof is not complete if you stop here, since you want to prove that those are equivalent definitions for homeomorphism. You have just proved the following:
$$[f\colon X\to Y\ \text{is a homeomorphism}] \implies [U \text{ is open in } X \iff f(U) \text{ is open in } Y]$$
Now, you would have to prove that the converse is also true. As an important remark (you didn't state this clearly), the bijectivity of $f$ has to be assumed in both directions, i.e., the two statements are equivalent when previously assuming that $f$ is a bijection. That means that you have to prove that:
$$[f:X\to Y \text{ is bijective and }U \text{ is open in } X \iff f(U) \text{ is open in } Y]\implies[f\colon X\to Y\ \text{is a homeomorphism}]$$
I will leave this proof up to you, since you already have all the tools you need. If you get stuck, don't hesitate to write a comment.
