continuity of a map into a one-point compactification

Question

Let $(E,\tau)$ be a topological space, $\Delta\not\in E$, $E^\ast:=E\cup\{\Delta\}$ and $$\tau^\ast:=\tau\cup\{E^\ast\setminus B:B\subseteq E\text{ is }\tau\text{-closed and }\tau\text{-compact}\}.$$ Note that, by definition, $\{\Delta\}$ is $\tau^\ast$-closed and $E^\ast$ is $\tau^\ast$-compact.

Assume $(\Omega,\mathcal A)$ is another topological space and $f:\Omega\to E^\ast$. Does $$f^{-1}(B)\in\mathcal A\;\;\;\text{for all }B\in\tau\tag1$$ already imply that $f$ is $(\mathcal A,\tau^\ast)$-continuous; i.e. $$f^{-1}(B)\in\mathcal A\;\;\;\text{for all }B\in\tau^\ast\tag2?$$

EDIT: I'm particularly interested in the case $\Omega=[0,\infty)$ equipped with the Euclidean topology.

Answer 1

In general it is not true. Let $\Omega = E^*$ and define a topology on $\Omega$ by $\mathcal A = \tau \cup \{E^*\}$. Then $E^*$ is the only open set containing $\Delta$.

Define $f : \Omega \to E^*, f(x) = x$. Then $f^{-1}(B) = B \in \tau \subset \mathcal A$ for $B \in \tau$.

Now assume that $E$ has a non-empty closed compact subset $B$. Then $E^* \setminus B \notin \mathcal A$. Since $f^{-1}(E^* \setminus B) = E^* \setminus B$, $f$ is not continuous.

Answer 2

What about when $E=\{0,1\}$ with discrete topology and $E^*=E\cup \{2\}$ . I have called $\Delta$ as $2$ here for easier notation.

Then by definition $\tau^*$ is the discrete topology on $\{0,1,2\}$ .

Now consider the any function from $f:(\Omega,\mathcal{A})\to (E^{*},\tau^*)$ .

Does it not require $f^{-1}(\{2\})$ to be open ?

For example when $(\Omega,\mathcal{A})=\bigg(\{a,b\},\bigg\{\{a,b\},\{a\},\phi\bigg\}\bigg)$ .

and $f(a)=1,f(b)=2$ . Then $f^{-1}(B)$ is open for all $B\in\tau$ .

But $f^{-1}(\{2\}\in\tau^{*})=\{b\}$ is not open in $\mathcal{A}$.

Edit:- If $E$ can be anything and $\Omega=[0,\infty)$ with usual(subspace) topology, then you can again use the abve example. Say $f:[0,\infty)\to E^{*}$ be $f(0)=2$ and $f(x)=1\,,\forall x>0$ . Then $f^{-1}(\{2\})=\{0\}$ is not open in $[0,\infty)$ . But $f^{-1}(\{0\})=\phi$ , $f^{-1}(\{1\})=f^{-1}(\{0,1\})=(0,\infty)$ is open.