Let 𝑋 and 𝑌 be two topological spaces. A continuous map 𝑓 ∶ 𝑋 → 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 ∈ 𝑋 ∶ 𝑓(𝑥) ∈ 𝐾}.

Consider ℝ with the usual topology. If ℝ ∖ {0} has the subspace topology induced from ℝ and ℝ × ℝ has the product topology, then which of the following maps is proper? 

Concept -

(i) A map 𝑓 : 𝑋 → 𝑌 is said to be proper if the preimage of any compact subset of 𝑌 is compact in 𝑋

 (ii) A set is compact if and only if it is closed and bounded. 

Explanation -

Let's consider 𝑓 : ℝ × ℝ → ℝ × ℝ defined by 𝑓(𝑥, 𝑦) = (𝑥 + 𝑦, 𝑦).

Let's see if the preimage of any compact subset of ℝ × ℝ is compact in ℝ × ℝ under this function.

First, consider a compact subset 𝐾 of ℝ × ℝ. Since ℝ × ℝ with the standard topology is a metric space, by the Heine-Borel Theorem, a set is compact if and only if it is closed and bounded. This means 𝐾 is a closed and bounded subset in ℝ × ℝ.

Take an arbitrary point (𝑥, 𝑦) in the preimage of 𝐾 under 𝑓, i.e., 𝑓^-1(𝐾).

This means 𝑓(𝑥, 𝑦) ∈ 𝐾. Since 𝐾 is closed in ℝ × ℝ, 𝑓(𝑥, 𝑦) should be a limit point of 𝐾.

Meanwhile, since 𝑓 is continuous, this implies (𝑥, 𝑦) is a limit point of  𝑓-1(𝐾) and  𝑓-1(𝐾) is closed in ℝ × ℝ.

Also, since 𝐾 is bounded, it can be enclosed within a sufficiently large square [-𝑎, 𝑎] x [-𝑏, 𝑏].

Any (𝑥, 𝑦) ∈  𝑓-1(𝐾) = {𝑥 ∈ 𝑋 ∶ 𝑓(𝑥) ∈ 𝐾} will also be included in the square with a larger side length, guaranteeing boundedness.

Hence,  𝑓-1(𝐾). is also bounded in ℝ × ℝ.

Thus, the preimage of any compact subset 𝐾 of ℝ × ℝ is compact in ℝ × ℝ, which implies the map 𝑓(𝑥, 𝑦) = (𝑥 + 𝑦, 𝑦) is a proper map.

Hence option (2) is correct.