Need some hint about proving that if $X\cong Y$ and $X$ is $T_0$ then $Y$ is $T_0$ Im following the book Topology without tears of Morris (my first serious attempt to understand this area of modern maths as an amateur) and Im having a lot of trouble with proofs (proofs are a very interesting puzzle to me now but sometimes are dark places that I dont know how to fill when Im writing one, this is the case for this question).
My attempt by now: by the definition of $T_0$-space we can get some $x_1,x_2\in X$ and $U\in\tau_1$ (we name the two topological spaces as $(X,\tau_1)$ and $(Y,\tau_2)$ where $x_1\in U \land x_2\notin U$ (or viceversa).
If we define the homeomorphism $f:X\to Y$ and $f(x_1)=y_1$, $f(x_2)=y_2$ and $f(U)=V$ where $y_1, y_2\in Y$ and $V\in\tau_2$ then I can "see" that $y_1\in V$ and $y_2\notin V$, so $Y$ is $T_0$ too... but this is not a formal proof, I dont know what I must write to make a valid proof.
Any hint or strategy would be very welcome. Thank you in advance.
 A: Suppose $f : X \to Y$ is a homeomorphism and $X$ is $T_0$. Let $y_1, y_2 \in Y$ be distinct points. Then $x_1 = f^{-1}(y_1)$ and $x_2 = f^{-1}(y_2)$ are distinct points of $X$, so since $X$ is $T_0$, there exists an open subset $U$ of $X$ such that, without loss of generality, $x_1 \in U$ and $x_2 \notin U$. Let $V = f(U)$. Since $f$ is a homeomorphism, $V$ is an open subset of $Y$, and since $x_1 \in U$ and $x_2 \notin U$, it follows that $y_1 \in V$ and $y_2 \notin V$. Thus, $Y$ is $T_0$.
A: What you wrote down is not a proof, since you got the order in your logical argument wrong. Before you write a proof, it's always good to write down what you know and what statement you want to prove:
We know:


*

*We have a homeomorphism $f$ between $X$ and $Y$.

*$X$ is $T_0$.


We want to prove:


*

*For all elements $y_1\neq y_2$ in $Y$ there exists an open set $V$ in $Y$ such that
either $y_1\in V$ and $y_2\notin V$ or vice versa.


So, given two elements $y_1\neq y_2$ in $Y$ you should try to find this set $V$. A natural way to proceed is to use the homeomorphism $f$ and look at $x_1=f^{-1}(y_1)$ and $x_2=f^{-1}(y_2)$. Then you can use the second fact that you know, that $X$ is $T_0$ for the two elements $x_1,x_2$. I will leave the rest for you to think about (or have a look at the other answer). 
A: When $T$ is a topology on $X$ ,and when $T'$ is a topology on $ X'$ ,and when they are homeomorphic, this means there is a continuous bijection $f:X\to X'$ for which $f^{-1}$ is also a continuous bijection. Then any property of $(X,T)$ that is stated entirely in topological terms is also a property of $(X',T').$ This is because $f$ induces a bijection $f^*:T\to T'$ where $f^*(U)=\{(f(x):x\in U\}.$ In particular, $x\in U\in T\iff f(x)\in f^*(U)\in T',$ and $(y\in X\backslash U\land U\in T) \iff (f(y)\in X'\backslash f^*(U)\land f^*(U)\in T').$
