Bijection Between Two Topologies: Continuity, Coarseness, And Fineness > Let $f: (X, \mathcal{T}_2) \rightarrow (X, \mathcal{T}_1)$ be a bijection between two topological spaces. Prove that $f$ is continuous if and only if $\mathcal{T}_1 \subset \mathcal{T}_2$ Here is my proof for the backward direction: Suppose $\mathcal{T}_1 \subset \mathcal{T}_2$ but $f$ is not continuous. Then there is a $U \in \mathcal{T}_1$ such that $f^{-1}(U) \notin \mathcal{T}_2$. Then since $f$ is bijective, $U \in \mathcal{T}_1 \subset f(\mathcal{T}_2)$, which is a contradiction. For the forward direction, suppose $f$ is continuous but $\mathcal{T}_1 \not\subset \mathcal{T}_2$. Then there is a $U \in \mathcal{T}_1$ such that $U \notin \mathcal{T}_2$. By the continuity of $f$, $f^{-1}(U) \in \mathcal{T}_2$, and here is where I'm not sure how to proceed.

Unfortunately neither implication is true. Let

$$X=\\{1,2\\}$$ $$\mathcal{T}_1=\\{\emptyset, X, \\{1\\}\\}$$ $$\mathcal{T}_2=\\{\emptyset, X, \\{2\\}\\}$$

Now let $f:X\to X$ be given by $f(1)=2$ and $f(2)=1$. You can easily check that $f$ is continuous (in fact it is a homeomorphism) but neither $\mathcal{T}_1\subseteq \mathcal{T}_2$ nor $\mathcal{T}_2\subseteq \mathcal{T}_1$.

* * *

On the other hand let

$$X=\\{1,2\\}$$ $$\mathcal{T}_1=\\{\emptyset, X, \\{1\\}\\}$$ $$\mathcal{T}_2=\\{\emptyset, X, \\{1\\}\\}$$

This time $\mathcal{T}_1\subseteq \mathcal{T}_2$ (they are even equal). Again consider $f(1)=2$ and $f(2)=1$ and note that it is not continuous.

* * *

The statement is not true for general bijection $f$. It is true however for the identity map $f(x)=x$.