Continuity in quotient topology
In quotient topology, a surjective function \( f: X \to A \) is continuous by construction, because a set \( V \subseteq A \) is open if and only if its preimage \( f^{-1}(V) \) is open in \( X \).
Consider a topological space \( X \) and a surjective function \( f: X \to A \), where \( A \) is any set, not necessarily a subset of \(X\).
The quotient topology on \( A \) ensures that the function \( f \) is continuous.
This works because in quotient topology, a set \( V \subseteq A \) is defined as open in \( A \) (relative to the quotient topology) if and only if the preimage of \( V \) under \( f \), i.e., \( f^{-1}(V) \), is open in \( X \).
Thus, this definition inherently satisfies the condition for \( f \) to be continuous.
Note. Thanks to this definition, the function \( f \) is automatically continuous, as the quotient topology is defined in terms of the preimages of open sets.
A practical example
Let’s consider a space \( X = \{a, b, c\} \), a set containing three points.
Now, define a surjective function \( f: X \to A \), where \( A = \{1, 2\} \), as follows:
- \( f(a) = f(b) = 1 \)
- \( f(c) = 2 \).
This function "groups" \( a \) and \( b \) together as the point \( 1 \) in \( A \).
In the quotient topology, a set \( V \subseteq A \) is open if its preimage under \( f \), i.e., \( f^{-1}(V) \), is open in \( X \).
For example, if we take \( V = \{1\} \subseteq A \), the preimage of \( V \) is \( f^{-1}(\{1\}) = \{a, b\} \). If \( \{a, b\} \) is open in \( X \), then \( V \) is open in \( A \).
Thus, the open sets in \( A \) are: \( \emptyset \), \( \{1, 2\} \), and \( \{2\} \).
- \( f^{-1}(\emptyset) = \emptyset \), which is the empty set and, being open in every topology, is also open in \( X \)
- \( f^{-1}(\{1, 2\}) = \{a, b\} \cup \{c\} = \{a, b, c\} \), which is the entire set \( X \), so it is open in \( X \)
- \( f^{-1}(\{1\}) = \{a\} \cup \{b\} = \{a, b\} \), which is open in \( X \)
- \( f^{-1}(\{2\}) = \{c\} \), which is open in \( X \)
This means that, in essence, the quotient topology on \( A \) makes the function \( f \) continuous, because whenever a set is open in \( A \), its preimage in \( X \) is also open.
Therefore, it is the very definition of quotient topology on \( A \) that ensures the continuity of the function \( f \).
And so on.
