TITLE: Countinuity of the identity map between different topologies
QUESTION [5 upvotes]: Could you please verify this (rather simple) proof? I'm a bit new to this kind of reasoning.
The question is:
Let $\mathcal{T}_1$, $\mathcal{T}_2$ be two topologies on some set $X$, when is the identity map $id:X \to X$, $id(x) = x$ continuous?
Here is my reasoning: 
If $\mathcal{T}_1 = \mathcal{T}_2$, then for any $S \in \mathcal{T}_2$, $id^{-1} (S) = S \in \mathcal{T}_1$ so then it's continuous.
If $\mathcal{T}_2 \subseteq \mathcal{T}_1$, then it's exactly like above, then it's also continouos.
Now:
If  $\mathcal{T}_1 \subseteq \mathcal{T}_2$ then take some $x \in U \in \mathcal{T}_2 \setminus \mathcal{T}_1$, then $id^{-1}(U) \not\in \mathcal{T}_1$, so it's not continuous.
So the identity map is continuous if $\mathcal{T}_1$ is finer than $\mathcal{T}_2$. 
Is this OK?
Is there a better way to show this?
Any extra applicable theory?
Thanks in advance

REPLY [4 votes]: That's almost perfect. Rather, you've (almost) showed that it's continuous if and only if $\mathcal T_1$ is finer than $\mathcal T_2.$
You only need two cases: $\mathcal T_2\subseteq\mathcal T_1,$ and $\mathcal T_2\not\subseteq\mathcal T_1.$ In the former case, you'll reason as you did in your equality case. In the latter case, you'll reason as you did in your $\mathcal T_1\subseteq\mathcal T_2$ approach, but it will actually be viable reasoning. (Can you figure out why yours was not?)