A description/proof of that induced functional structure on continuous topological spaces map codomain is functional structure
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of functional structure on topological space.
- The reader knows a definition of induced functional structure on continuous topological spaces map codomain.
Target Context
- The reader will have a description and a proof of the proposition that the induced functional structure on the codomain of any continuous topological spaces map is a functional structure.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any topological spaces, \(T_1, T_2\), any functional structure, \(\{F_{T_1} (U)\}\), and any continuous map, \(\phi: T_1 \to T_2\), the induced functional structure, \(\{F_{T_2} (U) = \{f: U \to \mathbb{R}\vert f \text{ is continuous } \land f \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\}\}\), is a functional structure.
2: Proof
Let us prove that the "induced functional structure" satisfies the 1st condition: \(F_{T_2} (U)\) is a sub-algebra of the all-the-continuous functions algebra. \(F_{T_2} (U)\) is a subset of the all-the-continuous functions algebra, because \(f\) is continuous by the definition. Let us check that the subset is closed under addition. Let us suppose that \(f_i \in F_{T_2} (U)\). \(f_1 + f_2 \in F_{T_2} (U)\)? \(f_1 + f_2\) is continuous. \((f_1 + f_2) \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\)? \(f_i \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\). \(f_1 \circ \phi + f_2 \circ \phi \in F_{T_1} (\phi^{-1} (U))\), but \(f_1 \circ \phi + f_2 \circ \phi = (f_1 + f_2) \circ \phi\). Let us check that the subset is closed under scalar multiplications. Let us suppose that \(f \in F_{T_2} (U)\) and \(r \in \mathbb{R}\). \(r f \in F_{T_2} (U)\)? \(r f\) is continuous. \((r f) \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\)? \(f \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\). \(r (f \circ \phi) \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\), but \(r (f \circ \phi) = (r f) \circ \phi\). Let us check that the subset is closed under multiplications. Let us suppose that \(f_i \in F_{T_2} (U)\). \(f_1 f_2 \in F_{T_2} (U)\)? \(f_1 f_2\) is continuous. \((f_1 f_2) \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\)? \(f_i \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\). \((f_1 \circ \phi) (f_2 \circ \phi) \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\), but \((f_1 \circ \phi) (f_2 \circ \phi) = (f_1 f_2) \circ \phi\). For any \(f_i \in F_{T_2} (U)\), the left distributability, \((f_1 + f_2) f_3 = f_1 f_3 + f_2 f_3\), holds. For any \(f_i \in F_{T_2} (U)\), the right distributability, \(f_3 (f_1 + f_2) = f_3 f_1 + f_3 f_1\), holds. For any \(f_i \in F_{T_2} (U)\) and \(r_i \in \mathbb{R}\), the compatibility with scalars, \((r_1 f_1) (r_2 f_2) = (r_1 r_2) (f_1 f_2)\), holds.
Let us prove that the "induced functional structure" satisfies the 2nd condition: \(F_{T_2} (U)\) contains all the constant functions. For any constant function, \(f: U \to \mathbb{R}\), \(f \circ \phi\) is constant, so, \(f \circ \phi \in F_{T_1} (\phi^{-1} (U))\).
Let us prove that the "induced functional structure" satisfies the 3rd condition: for any open \(V \subseteq U\) and any \(f \in F_{T_2} (U)\), \(f\vert_V \in F_{T_2} (V)\). \(f\vert_V \circ \phi \vert_{\phi^{-1} (V)} \in F_{T_1} (\phi^{-1} (V))\)? \(f \circ \phi \in F_{T_1} (\phi^{-1} (U))\). \(\phi^{-1} (V) \subseteq \phi^{-1} (U)\). \((f \circ \phi)\vert_{\phi^{-1} (V)} \in F_{T_1} (\phi^{-1} (V))\), but \((f \circ \phi)\vert_{\phi^{-1} (V)} = f\vert_V \circ \phi \vert_{\phi^{-1} (V)}\), because for any \(p \in \phi^{-1} (V)\), \((f \circ \phi)\vert_{\phi^{-1} (V)} (p) = f\vert_V \circ \phi (p)\).
Let us prove that the "induced functional structure" satisfies the 4th condition: for any open cover \(U = \cup_\alpha U_\alpha\) and any continuous \(f: U \to \mathbb{R}\) such that \(f \vert_{U_\alpha} \in F_{T_2} (U_\alpha)\), \(f \in F_{T_2} (U)\). \(f \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\)? \(f\vert_{U_\alpha} \circ \phi \vert_{\phi^{-1} (U_\alpha)} \in F_{T_1} (\phi^{-1} (U_\alpha))\), but \(f\vert_{U_\alpha} \circ \phi \vert_{\phi^{-1} (U_\alpha)} = (f \circ \phi)\vert_{\phi^{-1} (U_\alpha)}\), so, \((f \circ \phi)\vert_{\phi^{-1} (U_\alpha)} \in F_{T_1} (\phi^{-1} (U_\alpha))\). \(\phi^{-1} (U) = \cup_\alpha \phi^{-1} (U_\alpha)\). So, \((f \circ \phi)\vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\), but \((f \circ \phi)\vert_{\phi^{-1} (U)} = f \circ \phi \vert_{\phi^{-1} (U)}\), so, \(f \circ \phi \vert_{\phi^{-1} (U)} \in F_{T_1} (\phi^{-1} (U))\).
3: Note
Just calling \(\{F_{T_2} (U) = \{f: U \to \mathbb{R}\vert f \text{ is continuous } \land f \circ \phi \in F_{T_1} (\phi^{-1} (U))\}\}\) "induced functional structure" does not guarantee that \(\{F_{T_2} (U)\}\) is a functional structure, which requires satisfying the 4 conditions.
