topological spaces and continuous function

prove or disprove

Let $(X, \mathscr{T}_1)$ and $(Y, \mathscr{T}_2)$ be topological spaces and suppose that $f : X \to Y$ is a function that is $\mathscr{T}_1$ − $\mathscr{T}_2$ continuous. If f is one to one and $\mathscr{T}_2$ is the discrete topology on Y then $\mathscr{T}_1$ IS THE DISCRETE TOPOLOGY ON X.

I think,it is False statement since f is one to one but I do not know how can I come with counterexample.

• The discrete topology can be characterised as the topology in which every point is an open set. Can you see how to prove it now? – Joppy Apr 27 '17 at 2:50

Take a point $y \in I_{2}$, $y$ is open by the definition of discrete topology.

So, if $f$ is continuos, then $f^{1}(y) = {x}$ is an open set of $I_{1}$.

Now use $f$ is one to one.

• I know f is one to one that f(x1)=f(x2) implies x1=x2 but I do not know how can I apply it here could you help me please – rian asd Apr 27 '17 at 4:03
• @rianasd: What is the preimage of the set $\{f(x)\}$? – celtschk Apr 27 '17 at 4:15
• is it {y} ?.... – rian asd Apr 27 '17 at 4:17
• @rianasd: No, the preimage is a subset of $X$. – celtschk Apr 27 '17 at 4:18
• sorry, I am not sure what is it – rian asd Apr 27 '17 at 4:20

Suppose that $x \in X$. Then

$$f^{-1}[\{f(x)\}] = \{p \in X: f(p) \in \{f(x)\} \} = \{p \in X: f(p) =f(x) \} = \{x\}$$ where the last equality is a restatement that $f$ is 1-1.

$\{f(x)\}$ is open in $(Y,\mathscr{T}_2)$, as it is discrete. So it's inverse image under the continuous $f$ lies in $\mathscr{T}_1$, hence $\{x\}$ is open. This holds for every $x \in X$. Conclude that $X$ is indeed discrete.

Let $U$ be a subset of $x$. Thus $f[U]$ is an open subset of $Y$.
As $f$ is injective $U = f^{-1}[f[U]]$ is open.