Task:Characteristic property of the subspace topology

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

Statement

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

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

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

Proof

$f:Y\rightarrow S$ is continuous $\implies$ $i_S\circ f:Y\rightarrow X$ is continuous

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

$i_S\circ f:Y\rightarrow X$ is continuous $\implies$ $f:Y\rightarrow S$ is continuous

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

This completes the proof.