K. Jänich, Topology. Chapter I: Fundamental Concepts. 5: Continuous Maps

Definition: Continuous Function Let \(X\) and \(Y\) be topological spaces and \(f : X \rightarrow Y\). Then \(f\) is continuous iff, for any open set \(V\) in \(Y\), we have that \(f^{-1}(V)\) is open in \(X\).

Proposition 1 For any toplogical space \(X\), the identity map \(1_X : X \rightarrow X \) is continuous.

Proof If \(V\) is open in \(X\) then \(1_X^{-1}(V) = V\) is open in \(X\). \(\Box\)

Proposition 2 Let \(X\), \(Y\) and \(Z\) be topological spaces. If \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) are continuous then \( g \circ f : X \rightarrow Z \) is continuous.

Proof Let \(V\) be open in \(Z\). Then \(g^{-1}(V)\) is open in \(Y\), and so \(f^{-1}(g^{-1}(V)) = (g \circ f)^{-1}(V) \) is open in \(X\). \(\Box\)

Proposition 2.5Let \(X\) and \(Y\) be topological spaces. Any constant function \(X \rightarrow Y\) is continuous.

Proof Let \(b \in Y\). Let \(f : X \rightarrow Y\) be the constant function with value \(b\). For any open set \(V\) in \(Y\), we have \(f^{-1}(V) = X\) if \(b \in V\) and \(\emptyset\) if \(b \notin V\). In either case, \(f^{-1}(V)\) is open. \(\Box\)

Proposition 3 Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). Then the following are equivalent.

1. \(f\) is continuous.
2. For any closed set \(C\) in \(Y\), we have \(f^{-1}(C)\) is closed in \(X\).
3. For any \(x \in X\) and any neighbourhood \(N\) of \(f(x)\), we have \(f^{-1}(N)\) is a neighbourhood of \(x\).
4. For any \(B \subseteq Y\), we have \( \overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B}) \)

Proof

\(1 \Rightarrow 2\) Assume \(f\) is continuous. Let \(C\) be a closed set in \(Y\). Then \(Y - C\) is open, hence \(f^{-1}(Y - C) = X - f^{-1}(C) \) is open in \(X\), hence \(f^{-1}(C)\) is closed in \(X\).

\(2 \Rightarrow 1\) Similar.

\(1 \Rightarrow 3\) Assume \(f\) is continous. Let \(x \in X\). Let \(N\) be a neighbourhood of \(f(x)\). Pick an open set \(V\) such that \(f(x) \in V \subseteq N\). Then \(f^{-1}(V)\) is open and \(x \in f^{-1}(V) \subseteq f^{-1}(N)\), so \(f^{-1}(N)\) is a neighbourhood of \(x\) as required.

\(3 \Rightarrow 1\) Assume 3 holds. Let \(V\) be an open set in \(Y\). We shall prove that \(f^{-1}(V)\) is a neighbourhood of each of its points, which means it is open. So let \(x \in f^{-1}(V)\). Then \(f(x) \in V\), so \(V\) is a neighbourhood of \(f(x)\), and so \(f^{-1}(V)\) is a neighbourhood of \(x\) as required.

\(2 \Rightarrow 4\) Assume 2 holds. Let \(B \subseteq Y\). Then \(\overline{B}\) is closed, and so \(f^{-1}(\overline{B})\) is closed. We also have \(B \subseteq \overline{B}\), so \(f^{-1}(B) \subseteq f^{-1}(\overline{B})\). Thus \(f^{-1}(\overline{B})\) is a closed set that includes \(f^{-1}(B)\), and so \(\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})\).

\(4 \Rightarrow 2\) Assume 4 holds. Let \(C\) be a closed set in \(Y\). We have \(\overline{f^{-1}(C)} \subseteq f^{-1}(\overline{C}) = f^{-1}(C)\) and so \(\overline{f^{-1}(C)} = f^{-1}(C)\), which means \(f^{-1}(C)\) is closed. \(\Box\)

Proposition 5 Let \(X\) and \(Y\) be metric spaces and \(f : X \rightarrow Y \). Then \(f\) is continuous if and only if, for all \(x \in X\) and \(\epsilon > 0\), there exists \(\delta > 0\) such that, for all \(x' \in X\), if \(d(x,x') < \delta\) then \(d(f(x),f(x')) < \epsilon\).

ProofAssume \(f\) is continuous. Let \(x \in X\) and \(\epsilon > 0\). Then \(f^{-1}(B_\epsilon(f(x)))\) is a neighbourhood of \(x\). Pick \(\delta > 0\) such that \(B_\delta(x) \subseteq f^{-1}(B_\epsilon(f(x)))\). This says exactly that, for all \(x' \in X\), if \(d(x,x') < \delta\) then \(d(f(x),f(x')) < \epsilon\).

Conversely, assume that, for all \(x \in X\) and \(\epsilon > 0\), there exists \(\delta > 0\) such that, for all \(x' \in X\), if \(d(x,x') < \delta\) then \(d(f(x),f(x')) < \epsilon\). Let \(x \in X\) and \(N\) be a neighbourhood of \(f(x)\); we shall prove that \(f^{-1}(N)\) is a neighbourhood of \(x\). Pick \( \epsilon > 0\) such that \(B_\epsilon(f(x)) \subseteq N\). Pick \(\delta > 0\) such that, for all \(x' \in X\), if \(d(x,x') < \delta\) then \(d(f(x),f(x')) < \epsilon\). Then \(B_\delta(x) \subseteq f^{-1}(B_\epsilon(f(x))) \subseteq f^{-1}(N) \), and so \(f^{-1}(N)\) is a neighbourhood of \(x\) as required. \(\Box\).

Proposition 6 Let \(X\) be a topological space and \(Y\) be a subspace of \(X\). For any topological space \(Z\) and continuous function \(f : X \rightarrow Z\), the restriction \(f \restriction Y : Y \rightarrow Z\) in continuous.

Proof Give \(Y\) the subspace topology. Let \(f : X \rightarrow Z\) be continuous; we prove \(f \restriction Y : Y \rightarrow Z\) is continuous. Let \(V\) be an open set in \(Z\). Then \(f^{-1}(V)\) is open in \(X\), and so \((f \restriction Y)^{-1}(V) = f^{-1}(V) \cap Y\) is open in \(Y\). \(\Box\)

Proposition 7 Let \(X\) and \(Y\) be topological spaces. The disjoint union topology on \(X + Y\) is the unique topology such that, for any topological space \(Z\) and function \(f : X + Y \rightarrow Z\), we have \(f\) is continuous if and only if \(f \restriction X : X \rightarrow Z\) and \(f \restriction Y : Y \rightarrow Z\) are continuous.

ProofGive \(X + Y\) the disjoint union topology. Let \(f : X + Y \rightarrow Z\).

Assume \(f\) is continuous. Let \(V\) be open in \(Z\). Then \(f^{-1}(V)\) is open in \(X + Y\) and so \(f^{-1}(V) = U + W\) for some \(U\) open in \(X\) and \(W\) open in \(Y\). Hence \((f \restriction X)^{-1}(V) = U\) is open in \(X\) and \((f \restriction Y)^{-1}(V) = W\) is open in \(Y\).

Conversely, assume that \(f \restriction X\) and \(f \restriction Y\) are continuous. Let \(V\) be open in \(Z\). Then \((f \restriction X)^{-1}(V)\) is open in \(X\) and \((f \restriction Y)^{-1}(V)\) is open in \(Y\), and so \(f^{-1}(V) = (f \restriction X)^{-1}(V) + (f \restriction Y)^{-1}(V)\) is open in \(X + Y\).

Now, let \(\mathcal{O}\) be any topology on \(X + Y\) such that, for any topological space \(Z\) and function \( f : X + Y \rightarrow Z\), we have \(f\) is continuous if and only if \(f \restriction X\) and \(f \restriction Y\) are continuous.

The inclusions \(\kappa_X : X \rightarrow X + Y \) and \(\kappa_Y : Y \rightarrow X + Y\) are continuous, because they are the restrictions of \(1_{X+Y}\). Hence, for any \(U \in \mathcal{O}\), we have that \(U \cap X\) and \(U \cap Y\) are open in \(X\) and \(Y\) respectively, and so \(U = V + W\) for some \(V\) open in \(X\) and \(W\) open in \(Y\).

Conversely, for any \(V\) open in \(X\) and \(W\) open in \(Y\), define \(f : X + Y \rightarrow 2\) by \(f(x) = 1\) if \(x \in V + W\), and \(f(x)=0\) otherwise. Give \(2\) the topology \(\{\emptyset, \{1\}, \{0,1\}\}\). Then \(f \restriction X\) and \(f \restriction Y\) are continuous, hence \(f\) is continuous, and so \(f^{-1}(\{1\}) = V + W \) is open. \(\Box\)

Proposition 8Let \(X\) and \(Y\) be topological spaces. The product topology on \(X \times Y\) is the unique topology such that, for any topological space \(Z\) and function \(f : Z \rightarrow X \times Y\), we have that \(f\) is continuous if and only if \( \pi_1 \circ f : Z \rightarrow X\) and \(\pi_2 \circ f : Z \rightarrow Y\) are continuous.

Proof Give \(X \times Y\) the product topology. Let \(Z\) be a topological space and \(f : Z \rightarrow X \times Y\).

If \(f\) is continuous then \(\pi_1 \circ f\) and \(\pi_2 \circ f\) are continuous by Proposition 2.

Conversely, assume \(\pi_1 \circ f\) and \(\pi_2 \circ f\) are continuous. Let \(z \in Z\) and \(f(z) = (x,y)\). Let \(N\) be a neighbourhood of \(f(z)\). Pick neighbourhoods \(M_1\) of \(x\) and \(M_2\) of \(y\) such that \(M_1 \times M_2 \subseteq N\). Then \(f^{-1}(\pi_1^{-1}(M_1))\) and \(f^{-1}(\pi_2^{-1}(M_2))\) are neighbourhoods of \(z\), hence so is \(f^{-1}(\pi_1^{-1}(M_1)) \cap f^{-1}(\pi_2^{-1}(M_2)) = f^{-1}(M_1 \times M_2) \subseteq f^{-1}(N)\). Thus, \(f\) is continuous.

Now, let \(\mathcal{O}\) be any topology on \(X \times Y\) and assume that, for every topological space \(Z\) and function \(f : Z \rightarrow X \times Y\), we have that \(f\) is continuous if and only if \(\pi_1 \circ f\) and \(\pi_2 \circ f\) are continuous. Let \(\mathcal{P}\) be the product topology on \(X \times Y\).

We have \(\pi_1 : (X \times Y, \mathcal{O}) \rightarrow X\) and \(\pi_2 : (X \times Y, \mathcal{O}) \rightarrow Y\) are continuous (by taking \(f = 1_{X \times Y}\)). Therefore, from above, \( 1_{X \times Y} Z (X \times Y, \mathcal{O}) \rightarrow (X \times Y, \mathcal{P})\) is continuous. Hence \(\mathcal{P} \subseteq \mathcal{O}\).

We also have \(\pi_1 : (X \times Y, \mathcal{P}) \rightarrow X\) and \(\pi_2 : (X \times Y, \mathcal{P}) \rightarrow Y\) are continuous (by taking \(f = 1_{X \times Y}\)). Therefore, from above, \( 1_{X \times Y} Z (X \times Y, \mathcal{P}) \rightarrow (X \times Y, \mathcal{O})\) is continuous. Hence \(\mathcal{O} \subseteq \mathcal{P}\) and so \(\mathcal{O} = \mathcal{P}\). \(\Box\)

Definition 9 Let \(X\) and \(Y\) be topological spaces. A homeomorphism between \(X\) and \(Y\) is a bijection \(f : X \rightarrow Y\) such that both \(f\) and \(f^{-1}\) are continuous. We write \(f : X \cong Y\) iff \(f\) is a homeomorphism between \(X\) and \(Y\). We say spaces \(X\) and \(Y\) are homeomorphic, and write \(X \cong Y\), iff there exists a homeomorphism between them.