Identity function is a homeomorphism iff topologies are equal Let  $\tau_1$  and $\tau_2$ are two topologies on $X$
Then the function $f: (X,\tau_1) \to (X,\tau_2)$ defined by $f(x)=x$ is a homeomorphism if and only if $\tau_1=\tau_2$
I have tried many many unnecessary properties and I am in a jam now. I need exact and clear two directioned proof. Can someone illuminate me or direct me to exact proof? (Without using f is a homeomorphism iff f is open and continuous)
 A: Assume the identity map, $f:(X,\tau_1)\to(X,\tau_2)$ is a homeomorphism. Let $U\in\tau_1$, then by the continuity of $f^{-1}$, $(f^{-1})^{-1}(U)$ is open in $\tau_2$, but $(f^{-1})^{-1}(U) = f(U) = U$, so $U\in\tau_2$, hence $\tau_1\supseteq \tau_2$. Now let $U\in\tau_2$, then by the continuity of $f$, $f^{-1}(U)$ is open in $\tau_1$, but since the inverse of the identity map is itself (see https://proofwiki.org/wiki/Inverse_of_Identity_Mapping), $f^{-1}(U) = f(U) = U$, so $U\in\tau_1$, hence $\tau_2\supseteq \tau_1$. Therefore $\tau_1 = \tau_2$.
Now assume $\tau_1 = \tau_2$, and let $f:(X,\tau_1)\to(X,\tau_2)$ be the identity map. Then for every $U\in\tau_2$, $f^{-1}(U) = U$ and $U\in\tau_1$, hence $f$ is continuous. For every $U\in\tau_1$, $(f^{-1})^{-1}(U) = f(U) = U$ and $U\in\tau_2$, hence $f^{-1}$ is continuous. Additionally due to the definition of the identity map, it is a bijection (see https://proofwiki.org/wiki/Identity_Mapping_is_Bijection). As a result $f$ is a homeomorphism.
A: *

*If $\tau_1 = \tau_2$, then $f$ is definitely a homeomorphism, because it is bijective, because it is open (the image of every open set is itself an open set), and because it is continuous (the inverse image of every open set is itself open.)

*If the identity function $f$ is a homeomorphism, then it is continuous and open. 
Because $f$ is open: if $A$ is an open set in $\tau_1$, then $f(A)$ is an open set in $\tau_2$. But $f(A) = A$ because $f$ is the identity, so this shows that $A$ is open in $\tau_2$ if $A$ is open in $\tau_1$.
Because $f$ is continuous: if $B$ is an open set in $\tau_2$, then $f^{-1}(B)$ is an open set in $\tau_1$. But $f^{-1}(B) = B$ because $f^{-1}$ is the identity, so this shows that $B$ is open in $\tau_1$ if $B$ is open in $\tau_2$.
This shows that $U$ is open in $\tau_1$ if and only if $U$ is open in $\tau_2$; hence $\tau_1 = \tau_2$.
