Discussion
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?
Answer: B
The proper map is \(f(x,y) = (x+y, y)\) on \(\mathbb{R} \times \mathbb{R}\).
This f is linear with inverse \((u, v) \mapsto (u - v, v)\), so it is a homeomorphism of the plane. The preimage of a compact set is its image under the continuous inverse, which is compact. So f is proper.
The others fail. The inclusion of \(\mathbb{R} \setminus \{0\}\) into \(\mathbb{R}\) pulls back [-1, 1] to \([-1,0) \cup (0,1]\), not compact. The projection (x, y) → x pulls back the point {0} to the whole y-axis, unbounded. And \(x^{2}-y^{2}\) pulls back {0} to the two lines y = ±x, also unbounded.
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