Question

Let X and Y be two topological spaces. A continuous map \(f: X \to Y\) is said to be proper if \(f^{-1}(K)\) is compact in X for every compact subset K of Y, where \(f^{-1}(K)\) is defined by \(f^{-1}(K) = \{x \in X : f(x) \in K\}\).
Consider \(\mathbb{R}\) with the usual topology. If \(\mathbb{R} \setminus \{0\}\) has the subspace topology induced from \(\mathbb{R}\) and \(\mathbb{R} \times \mathbb{R}\) has the product topology, then which of the following maps is proper?

• A. \(f: \mathbb{R} \setminus \{0\} \to \mathbb{R}\) defined by \(f(x) = x\)
• B. \(f: \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) defined by \(f(x,y) = (x+y, y)\)
• C. \(f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}\) defined by \(f(x,y) = x\)
• D. \(f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}\) defined by \(f(x,y) = x^2 - y^2\)

Hint:

To quickly test if a map \(f: \mathbb{R}^n \to \mathbb{R}^m\) is proper, check the preimage of a simple compact set, like a single point \(\{p\}\). If \(f^{-1}(\{p\})\) is unbounded, the map is not proper. Projections and maps that reduce dimension often fail this test. Homeomorphisms on \(\mathbb{R}^n\) are often proper maps.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A continuous map \(f: X \to Y\) is proper if the preimage of any compact set in Y is a compact set in X. In Euclidean spaces \(\mathbb{R}^n\), a set is compact if and only if it is closed and bounded (Heine-Borel Theorem). We need to check this condition for each given map.
Step 3: Detailed Explanation:
(A) Let \(f(x) = x\). Consider the compact set \(K = [1, 2]\) in \(\mathbb{R}\). The preimage is \(f^{-1}(K) = \{x \in \mathbb{R} \setminus \{0\} : x \in [1, 2]\} = [1, 2]\). This is a compact set. However, consider \(K = [-1, 1]\), which is compact in \(\mathbb{R}\).
The preimage is \(f^{-1}(K) = [-1, 0) \cup (0, 1]\). This set is bounded, but it is not closed in \(\mathbb{R}\) (it is missing 0 and its closure is \([-1,1]\)).
Therefore, it is not compact. So, \(f\) is not proper.
(B) Let \(f(x,y) = (x+y, y)\). This is a linear transformation. We can write its inverse by letting \(u = x+y\) and \(v = y\). Then \(y=v\) and \(x=u-v\). So, \(f^{-1}(u,v) = (u-v, v)\).
The inverse function \(f^{-1}\) is also continuous. A continuous map between topological spaces that has a continuous inverse is a homeomorphism. Let \(K\) be a compact subset of \(\mathbb{R}^2\).
The preimage \(f^{-1}(K)\) is the image of the compact set \(K\) under the continuous map \(f^{-1}\). The continuous image of a compact set is compact. Therefore, \(f^{-1}(K)\) is compact for every compact \(K\). The map \(f\) is proper.
(C) Let \(f(x,y) = x\). This is the projection onto the x-axis. Consider the compact set \(K = \{0\}\) in \(\mathbb{R}\). The preimage is \(f^{-1}(\{0\}) = \{(x,y) \in \mathbb{R}^2 : x=0\}\). This is the y-axis. The y-axis is a closed set, but it is not bounded. Therefore, it is not compact. So, \(f\) is not proper.
(D) Let \(f(x,y) = x^2 - y^2\). Consider the compact set \(K = \{0\}\) in \(\mathbb{R}\). The preimage is \(f^{-1}(\{0\}) = \{(x,y) \in \mathbb{R}^2 : x^2 - y^2 = 0\}\). This corresponds to the set where \((x-y)(x+y) = 0\), which is the union of the two lines \(y=x\) and \(y=-x\). This set is closed but not bounded, so it is not compact. Therefore, \(f\) is not proper.
Step 4: Final Answer:
The only proper map is \(f(x,y) = (x+y, y)\).
Step 5: Why This is Correct:
The map in (B) is a homeomorphism of \(\mathbb{R}^2\) onto itself. The continuous image (under the inverse map) of any compact set is compact, which satisfies the definition of a proper map. The other maps fail because the preimages of certain compact sets (like a single point) are unbounded sets (lines), which are not compact.