Task:Characteristic property of the subspace topology

Stub grade: A

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Needs to be added to the "pending task" category (when that template is created) and also needs to have the proof reviewed

Statement

[ilmath]\xymatrix{ Y \ar[r]^f \ar[dr]_{i_S\circ f} & S \ar@{_{(}->}[d]^{i_S} \\ & X}[/ilmath]
Diagram

Suppose [ilmath](X,\mathcal{ J })[/ilmath] is a topological space and [ilmath]S\in\mathcal{P}(X)[/ilmath] is a set which we endow with the subspace topology, so becomes a topological space, [ilmath](S,\mathcal{ K })[/ilmath] say. Suppose that [ilmath](Y,\mathcal{ L })[/ilmath] is any topological space and [ilmath]f:Y\rightarrow S[/ilmath] is a map. Then:

• [ilmath]f:Y\rightarrow S[/ilmath] is continuous if and only if the map [ilmath]i_S\circ f:Y\rightarrow X[/ilmath] is continuous (where [ilmath]i_S:S\rightarrow X[/ilmath] is the canonical injection given by [ilmath]i_S:s\mapsto s[/ilmath])

Proof

[ilmath]f:Y\rightarrow S[/ilmath] is continuous [ilmath]\implies[/ilmath] [ilmath]i_S\circ f:Y\rightarrow X[/ilmath] is continuous

• Let [ilmath]U\in\mathcal{J} [/ilmath] be an open set in [ilmath]X[/ilmath]; we must show [ilmath](i_S\circ f)^{-1}(U)\in\mathcal{L} [/ilmath] (that [ilmath](i_S\circ f)^{-1}(U)[/ilmath] is open in [ilmath]Y[/ilmath])
  • Note that [ilmath](i_S\circ f)^{-1}(U)=(f^{-1}\circ i_S^{-1})(U)=f^{-1}(i_S^{-1}(U))[/ilmath] (using sufficient abusive of notation, as [ilmath]g^{-1} [/ilmath] for any map [ilmath]g[/ilmath] might not be a map itself)
  • Furthermore, [ilmath]i_S^{-1}(U)=U\cap S[/ilmath]
  • Thus we see [ilmath](i_S\circ f)^{-1}(U)=f^{-1}(U\cap S)[/ilmath]
  • By hypothesis, [ilmath]f:Y\rightarrow S[/ilmath] is continuous, so:
    • [ilmath]\forall V\in\mathcal{K}[f^{-1}(V)\in\mathcal{L}][/ilmath]
  • Thus the proof hinges on whether or not [ilmath]U\cap S\in\mathcal{K} [/ilmath]
    • By definition of [ilmath]\mathcal{K} [/ilmath] (the (subspace) topology on [ilmath]S[/ilmath]) we see:
      • [ilmath]A\in\mathcal{K}\iff\exists B\in\mathcal{J}[B\cap S=A][/ilmath]
    • Defining [ilmath]A:=U\cap S[/ilmath] we see, as [ilmath]U\in\mathcal{J} [/ilmath] that there does exist a [ilmath]B\in\mathcal{J} [/ilmath] such that [ilmath]B\cap S=A=U\cap S[/ilmath] - namely [ilmath]B:=U[/ilmath] itself.
  • We have shown [ilmath]U\cap S\in\mathcal{K} [/ilmath], thus by hypothesis of continuity of [ilmath]f:Y\rightarrow S[/ilmath] we see:
    • [ilmath]f^{-1}(U\cap S)\in\mathcal{L}[/ilmath] - that is to say that [ilmath]f^{-1}(U\cap S)[/ilmath] is open in [ilmath]Y[/ilmath].
  • Since [ilmath](i_S\circ f)^{-1}(U)=f^{-1}(U\cap S)\in\mathcal{L}[/ilmath] we see [ilmath](i_S\circ f)^{-1}(U)\in\mathcal{L}[/ilmath]
• As [ilmath]U\in\mathcal{J} [/ilmath] was arbitrary we have shown: [ilmath]\forall U\in\mathcal{J}[(i_S\circ f)^{-1}(U)\in\mathcal{L}][/ilmath] - that [ilmath](i_S\circ f)[/ilmath] is continuous, as required.

[ilmath]i_S\circ f:Y\rightarrow X[/ilmath] is continuous [ilmath]\implies[/ilmath] [ilmath]f:Y\rightarrow S[/ilmath] is continuous

• Let [ilmath]U\in\mathcal{K} [/ilmath] be given ([ilmath]U[/ilmath] is open in [ilmath]S[/ilmath]), we wish to show that [ilmath]f^{-1}(U)\in\mathcal{L} [/ilmath], that [ilmath]f^{-1}(U)[/ilmath] is open in [ilmath]Y[/ilmath].
  • Then, as [ilmath](S,\mathcal{ K })[/ilmath] is a subspace, there exists a [ilmath]V\in\mathcal{J} [/ilmath] ([ilmath]V[/ilmath] open in [ilmath]X[/ilmath]) such that [ilmath]U=V\cap S[/ilmath]
    • As [ilmath]i_S\circ f:Y\rightarrow X[/ilmath] is continuous:
      • [ilmath]\forall A\in\mathcal{J}[(i_S\circ f)^{-1}(A)\in\mathcal{L}][/ilmath]
    • So we see [ilmath](i_S\circ f)^{-1}(V)\in\mathcal{L} [/ilmath]
      • This means [ilmath]f^{-1}(i_S^{-1}(V))\in\mathcal{L} [/ilmath]
    • But [ilmath]i_S^{-1}(V)=V\cap S[/ilmath], so
      • We're really saying: [ilmath]f^{-1}(V\cap S)\in\mathcal{L}[/ilmath]
    • Recall: [ilmath]U=V\cap S[/ilmath], so:
    • [ilmath]f^{-1}(U)\in\mathcal{L}[/ilmath], as required.
• Since [ilmath]U\in\mathcal{K} [/ilmath] was arbitrary we have shown: [ilmath]\forall U\in\mathcal{K}[f^{-1}(U)\in\mathcal{L}][/ilmath], the very definition of [ilmath]f[/ilmath] being continuous.

This completes the proof.