Explanation of conditions when this map is open

Question

As an example of a continuous function that needs not map open sets to open sets, our professor exhibited the identity map:

$$ 1_{x}:(X, \mathcal{T}) \to (X, \mathcal{O})$$ $$ x \mapsto x$$

when $\mathcal{O}$ is a proper subset of $\mathcal{T}$, but I'm not sure I understand why.

Could someone please explain this to me?

If $x \in \mathcal{T}$ but is not in $\mathcal{O}$, it seems to me that it could be mapped to $\emptyset$, which is always open. Or would this then contradict the fact that it's the identity map, and thus, $x$ needs to be mapped to itself?

I'm possibly overthinking this, but humor me anyway.

Answer 1

$1_{x}$ is the identity map so in particular it maps every subset to itself.

A map is continuous if and only if the inverse image of any open set is open. If $U \in \mathcal{O}$ then since $\mathcal{O} \subset \mathcal{T}$ we must have $U \in \mathcal{T}$ so $1_x$ is continuous.

On the other hand since $\mathcal{O}$ is a proper subset of $\mathcal{T}$, there exists $T \in \mathcal{T} \setminus \mathcal{O}$.

Then $T$ is the inverse image under $1_x^{-1}$ of an open set, yet it fails to be open. So $1_x^{-1}$ is not continuous.

Answer 2

You are definitely overthinking it. Let $U\in\tau$ with $U\notin\mathcal{O}$. Then for every $x\in U$, $1(x)=x$, which means that $U\mapsto U$ (it is just the identity function after all). In the domain, $U$ is open while in the range $U$ is not open (it is specifically chosen to not be in the topology).

Answer 3

Note that for the identity function $1_X$ we have for any subsets $A,B \subseteq X$:

$$1_X[A] = \{1_X(x): x \in A\} = \{x: x \in A\} =A$$ and

$$(1_X)^{-1}[B] = \{x \in X: 1_X(x) \in B\}= \{x \in X: x \in B\} = B$$

So when we have $i_X: (X,\mathcal{T}_1) \to (X, \mathcal{T}_2)$ we have that $1_X$ is continuous iff $$\forall O \in \mathcal{T}_2 :(1_X)^{-1}[O] = O \in \mathcal{T}_1$$ so $i_X$ is continuous iff $\mathcal{T}_2 \subseteq \mathcal{T}_1$ and $i_X$ is open iff $$\forall O \in \mathcal{T}_1: i_X[O] = O \in \mathcal{T}_2$$ so $i_X$ is open iff $\mathcal{T_1} \subseteq \mathcal{T}_2$.

So if we choose (as your professor did) $\mathcal{O} \subseteq \mathcal{T}$ while $\mathcal{T} \nsubseteq \mathcal{O}$, we get a continuous and non-open map $i_X:(X,\mathcal{T}) \to (X, \mathcal{O})$.