9.5. Continuous Mappings¶
In most branches of pure mathematics we study what in category theory are called “objects” and “arrows”.
In linear algebra the objects are vector spaces and the arrows are linear transformations.
In group theory, the objects are groups and the arrows are homomorphisms.
In set theory, the objects are sets and arrows are functions.
In topology, the objects are topological spaces. The arrows are continuous mappings.
9.5.1. Continuous Mappings¶
Motivation
A function \(f: \mathbb{R} \rightarrow \mathbb{R}\) is called continuous if for each \(a \in \mathbb{R}\) and each positive real number \(\epsilon\), there exists a positive real number \(\delta\) such that \(|x - a| < \delta\) implies \(|f(x) - f(a)| < \epsilon\).
We need to generalize this notion for arbitrary topological spaces without the notion of absolute value or subtraction.
Restating: \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous if and only if for each \(a \in \mathbb{R}\) and each interval \((f(a) - \epsilon, f(a) + \epsilon)\) for \(\epsilon > 0\), there exists \(\delta > 0\) such that \(f(x) \in (f(a) - \epsilon, f(a) + \epsilon)\) for all \(x \in (a - \delta, a + \delta)\).
Lemma
Let \(f\) be a function mapping \(\mathbb{R}\) into itself. Then \(f\) is continuous if and only if for each \(a \in \mathbb{R}\) and each open set \(U \ni f(a)\), there exists an open set \(V\) containing \(a\) such that \(f(V) \subseteq U\).
Almost ready to generalize for topological spaces!
Lemma
Let \(f\) be a mapping from a topological space \((X, \mathcal{T})\) into a topological space \((Y, \mathcal{T}')\). Then the following two conditions are equivalent:
for each \(U \in \mathcal{T}', f^{-1}(U) \in \mathcal{T}\),
for each \(a \in X\) and each \(U \in \mathcal{T}'\) with \(f(a) \in U\), there exists a \(V \in \mathcal{T}\) such that \(a \in V\) and \(f(V) \subseteq U\).
\(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous if and only if for each open subset \(U\) of \(\mathbb{R}\), \(f^{-1}(U)\) is an open set.
Definition
Let \(f\) be a mapping from a topological space \((X, \mathcal{T})\) into a topological space \((Y, \mathcal{T}')\). Then \(f : (X, \mathcal{T}) \rightarrow (Y, \mathcal{T}')\) is said to be a continuous mapping if for each \(U \in \mathcal{T}'\), \(f^{-1}(U) \in \mathcal{T}\).
Example
Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x) = c\) a constant. Let \(U\) be any open set in \(\mathbb{R}\). Clearly \(f^{-1}(U) = \mathbb{R} \text{ if } c \in U\) and \(\phi \text{ if } c \notin U\). In both cases, \(f^{-1}(U)\) is open, hence \(f\) is a continuous mapping.
Proposition
Let \(f\) be a mapping of a topological space \((X, \mathcal{T})\) into a topological space \((Y, \mathcal{T}')\). Then \(f\) is continuous if and only if for each \(x \in X\) and each \(U \in \mathcal{T}' | f(x) \in U\), there exists a \(V \in \mathcal{T}\) such that \(x \in V\) and \(f(V) \subseteq U\).
Is continuity associative?
Proposition
Let \((X, \mathcal{T}_1)\), \((Y, \mathcal{T}_2)\) and \((Z, \mathcal{T}_3)\) be topological spaces. If \(f : (X, \mathcal{T}_1) \rightarrow (Y, \mathcal{T}_2)\) and \(g : (Y, \mathcal{T}_2) \rightarrow (Z, \mathcal{T}_3)\) are continuous mappings, then the composition function \(f \circ g : (X, \mathcal{T}_1)\rightarrow (Z, \mathcal{T}_3)\) is continuous.
The continuity can be described in terms of closed sets also.
Proposition
Let \((X, \mathcal{T}_1)\) and \((Y, \mathcal{T}_2)\) be topological spaces. Then \(f : (X, \mathcal{T}_1) \rightarrow (Y, \mathcal{T}_2)\) is continuous if and only if for every closed subset \(S\) of \(Y\), \(f^{-1}(S)\) is a closed subset of \(X\).
Are continuous mappings and homeomorphisms related?
Remark
If \(f : (X, \mathcal{T}_1) \rightarrow (Y, \mathcal{T}_2)\) is a homeomorphism then it is a continuous map. But every continous mapping doesn’t need to be a homeomorphism.
Which mappings are homeomorphisms?
Proposition
Let \((X, \mathcal{T}_1)\) and \((Y, \mathcal{T}_2)\) be topological spaces. Let \(f\) be a function from \(X\) into \(Y\). Then \(f\) is a homeomorphism if and only if
\(f\) is continuous
\(f\) is one-one and onto.
\(f^{-1}\) is continuous.
What about subspaces? If \(f\) is continuous on \(X\) is it continuous on its subspaces too?
Proposition
Let \((X, \mathcal{T}_1)\) and \((Y, \mathcal{T}_2)\) be topological spaces. And \(f : (X, \mathcal{T}_1) \rightarrow (Y, \mathcal{T}_2)\) be a continuous mapping. Let \(A\) be a subset of \(X\) and let \(\mathcal{T}_3\) be the induced topology on \(A\). Further, let \(g : (A, \mathcal{T}_3) \rightarrow (Y, \mathcal{T}_2)\) be a restriction of \(f\) to \(A\); that is, \(g(x) = f(x) \quad \forall x \in A\). Then \(g\) is continuous.
Definition
Let \(\mathcal{T}_1\) and \(\mathcal{T}_2\) be two topologies on \(X\). Then \(\mathcal{T}_1\) is said to be a finer topology than \(\mathcal{T}_2\) (and \(\mathcal{T}_2\) is said to be a coarser topology than \(\mathcal{T}_1\)) if \(\mathcal{T}_1 \supseteq \mathcal{T}_2\).
9.5.2. Intermediate Value Theorem¶
Is continuity and connectivity related?
Proposition
Let \((X, \mathcal{T}_1)\) and \((Y, \mathcal{T}_2)\) be topological spaces and \(f : (X, \mathcal{T}_1) \rightarrow (Y, \mathcal{T}_2)\) be surjective and continuous. If \((X, \mathcal{T}_1)\) is connected then \((Y, \mathcal{T}_2)\) is connected.
Any continuous image of a connected set is connected.
If \((X, \mathcal{T}_1)\) is connected and \((Y, \mathcal{T}_2)\) is disconnected, then there exists no mapping which is onto and continuous.
While there are infinite number of mappings from \(\mathbb{R}\) onto \(\mathbb{Q}\) or \(\mathbb{Z}\), none of them are continuous.
But the notion of connectivity is too broad. Can we do something better?
Definition
A topological space \((X, \mathcal{T})\) is said to be path connected (or pathwise connected) if for each pair of distinct points \(a,b \in X\) there exists a continuous mapping \(f:[0,1] \rightarrow (X, \mathcal{T})\), such that \(f(0) = a\) and \(f(1) = b\). The mapping \(f\) is said to be a path joining a to b.
Example
Every interval is path connected.
For each \(n \geq 1, \mathbb{R}^n\) is path connected.
Proposition
Every path connected space is connected.
Not every connected space is path connected!
Example
Clearly \(\mathbb{R}^2 \setminus \{\langle 0, 0 \rangle \}\) is path connected hence is connected. However, \(\mathbb{R} \setminus \{a\}\) for any \(a \in \mathbb{R}\) is disconnected. Hence \(\mathbb{R} \ncong \mathbb{R}^2\).
Theorem
Weierstrass Intermediate Value Theorem Let \(f: [a,b] \rightarrow \mathbb{R}\) be continuous and let \(f(a) \neq f(b)\). Then for every number \(p\) between \(f(a)\) and \(f(b)\) there is a point \(c \in [a,b]\) such that \(f(c) = p\).
Corollary
If \(f: [a,b] \rightarrow \mathbb{R}\) is continuous and such that \(f(a) > 0\) and \(f(b) < 0\), then there exists an \(x \in [a,b]\) such that \(f(x) = 0\).
Corollary
Fixed Point Theorem Let \(f\) be a continuous mapping of \([0,1]\) into \([0,1]\). Then there exists a \(z \in [0,1]\) such that \(f(z) = z\). The point \(z\) is called a fixed point.