Identity function is a homeomorphism iff topologies are equal

Question

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)

Answer 1

1. 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.)

2. 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 $\tau_1 = \tau_2$.