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Continuous Functions
Now we define the notion of continuous functions. Intuitively, one can think of this as a function preserving topological structure, just as a homomorphism does in an algebraic setting.
Definition 1 A function \(f:X\rightarrow Y\) between arbitrary topological spaces \(X\) and \(Y\) is continuous at \(x\in X\) if for any neighborhood \(V\) of \(f(x)\in Y\), there exists a neighborhood \(U\) of \(x\) such that \(f(U)\subseteq V\).
Since \(U\subseteq f^{-1}(f(U))\) holds for any function \(f:X\rightarrow Y\) between arbitrary sets \(X\), \(Y\) and any \(U\subseteq X\), if an arbitrary set \(V\subseteq Y\) satisfies \(f(U)\subseteq V\), then
\[U\subseteq f^{-1}(f(U))\subseteq f^{-1}(V)\]holds. Therefore, to prove that a function \(f:X\rightarrow Y\) between two topological spaces is continuous at \(x\in X\), it suffices to show that for any neighborhood \(V\) of \(f(x)\in Y\), \(f^{-1}(V)\) is a neighborhood of \(x\); more generally, it suffices to show that for a fixed local base \(\mathcal{B}(f(x))\) of \(f(x)\) and any \(V\in\mathcal{B}(f(x))\), we have \(f^{-1}(V)\in\mathcal{N}(x)\).
Proposition 2 Let \(f:X\rightarrow Y\) be a function between two topological spaces \(X\) and \(Y\) that is continuous at a point \(x\). If \(x\in\cl(A)\) for some \(A\subseteq X\), then \(f(x)\in\cl(A)\).
Proof
Let \(V\) be an arbitrary neighborhood of \(f(x)\in Y\). Then \(f^{-1}(V)\) is a neighborhood of \(x\), so \(f^{-1}(V)\cap A\neq\emptyset\) (§Interior, Closure, and Boundary of Sets, ⁋Proposition 6). If we let \(x'\in f^{-1}(V)\cap A\), then \(f(x')\in V\cap f(A)\). In particular, \(V\cap f(A)\neq\emptyset\), so applying §Interior, Closure, and Boundary of Sets, ⁋Proposition 6 again, we see that \(f(x)\in\cl(A)\).
Proposition 3 Let topological spaces \(X\), \(Y\), \(Z\) be given. If \(f:X\rightarrow Y\) is continuous at a point \(x\in X\) and \(g:Y\rightarrow Z\) is continuous at \(f(x)\), then their composition \(g\circ f\) is also continuous.
Proof
Let \(W\) be an arbitrary neighborhood of \((g\circ f)(x)\). Then since \(g\) is continuous at \(f(x)\), \(g^{-1}(W)\) is a neighborhood of \(f(x)\). Again, since \(f\) is continuous at \(x\), \(f^{-1}(g^{-1}(W))\) is a neighborhood of \(x\). ([Set Theory] §Operations of Binary Relations, ⁋Proposition 13)
If \(f\) is continuous at every point of \(X\), we call \(f\) a continuous function. The following theorem shows several equivalent ways to define continuous functions.
Theorem 4 For two topological spaces \(X\), \(Y\) and a function \(f:X\rightarrow Y\), the following are all equivalent.
- \(f\) is continuous.
- For any subset \(A\) of \(X\), \(f(\cl A)\subset\cl f(A)\) holds.
- For any closed set \(C\) of \(Y\), \(f^{-1}(C)\) is a closed set in \(X\).
- For any open set \(V\) of \(Y\), \(f^{-1}(V)\) is an open set in \(X\).
Proof
That the first condition implies the second is a consequence of Proposition 2.
Now assume the second condition and show the third. For any closed set \(C\) of \(Y\), the inclusion
\[f(\cl(f^{-1}(C))\subseteq \cl(f(f^{-1}(C))\subseteq\cl(C)=C\]holds, so from
\[\cl(f^{-1}(C))\subseteq f^{-1}(f(\cl(f^{-1}(C)))\subseteq f^{-1}(C)\]we see that \(f^{-1}(C)\) is a closed set. Since the identity \((f^{-1}(A))^c=f^{-1}(A^c)\) holds for any subset \(A\subseteq Y\), it is obvious that the fourth condition follows as well.
Thus it suffices to assume the fourth condition and prove the first. Choose \(x\in X\) arbitrarily, and let \(V\) be any neighborhood of \(f(x)\in Y\). Then there exists an
By Proposition 3, if two continuous functions \(f:X\rightarrow Y\), \(g:Y\rightarrow Z\) are given, we know that \(g\circ f\) is also continuous.
Let a continuous function \(f:X\rightarrow Y\) between two topological spaces \((X,\mathcal{T}_X)\), \((Y,\mathcal{T}_Y)\) be given. Then since \(f^{-1}(V)\in\mathcal{T}_X\) holds for any \(V\in\mathcal{T}_Y\), the formula
\[f^\mathcal{T}(V):=f^{-1}(V),\qquad V\in\mathcal{T}_Y\]well-defines a function \(f^\mathcal{T}:\mathcal{T}_Y\rightarrow\mathcal{T}_X\).
Now assume \(f\) is a bijection and consider two distinct elements \(V_1\), \(V_2\) of \(\mathcal{T}_Y\). Without loss of generality, if \(y\in V_1\setminus V_2\), then
\[f^{-1}(y)\in f^\mathcal{T}(V_1)\setminus f^\mathcal{T}(V_2)\]holds, so \(f^{\mathcal{T}}\) is injective.
Example 5 In general, \(f^{\mathcal{T}}\) need not be surjective. For example, let \(X_1\) be the space \(\mathbb{N}\) with the discrete topology \(\mathcal{T}_1\), and \(X_2\) be the space with the trivial topology \(\mathcal{T}_2\); then considering the identity function \(\id:\mathbb{N}\rightarrow\mathbb{N}\) as a set map, \(\id\) is a continuous bijection but the function
\[\id^\mathcal{T}:\mathcal{T}_2\rightarrow\mathcal{T}_1\]cannot be surjective. ([Set Theory] §Operations of Cardinals, ⁋Proposition 10)
However, if the inverse function \(f^{-1}\) of a bijection \(f\) is also continuous, then \((f^{-1})^\mathcal{T}:\mathcal{T}_X\rightarrow\mathcal{T}_Y\) is well-defined, and from the definition it is obvious that
\[f^\mathcal{T}\circ (f^{-1})^\mathcal{T}=\id_{\mathcal{T}_X},\qquad (f^{-1})^\mathcal{T}\circ f^\mathcal{T}=\id_{\mathcal{T}_Y}\]so \(f^\mathcal{T}\) is also bijective. We define such a situation as follows.
Definition 6 A continuous function \(f:X\rightarrow Y\) is a homeomorphism if there exists another continuous function \(g:Y\rightarrow X\) such that \(f\circ g=\id_Y\) and \(g\circ f=\id_X\).
That is, two topological spaces \(X\) and \(Y\) are homeomorphic means not only that there exists a bijection between them as sets, but also that this bijection acts on the topological structures of the two sets, i.e., on their open sets, in exactly the same way. An example of a continuous bijection that is not a homeomorphism is precisely Example 5 above.
References
[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.
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