Function that preserves compactness?

Question

Let S be a compact topological space, let T be a topological space, and let f be a function from S onto T . Which of the following conditions on f is the weakest sufficient condition to guarantee compactness of T : (a) f is a homeomorphism (b) f is continuous and injective (c) f is continuous (d) f is injective (e) f is bounded I think (c) is the weakest condition that guarantees compactness of T. I can construct a bounded function from a compact set to a non compact set, but I am not quite sure about (d). It means f is bijective (since it's given that f maps from S onto T), so I've been trying to find a bijective function that maps from a compact set, preferably [0,1], to a non compact set.

Answer 1

It is a well known result in topology that if S is compact and f is a continuous map then f(S) is compact. Since f is also assumed to be onto we have that f(S) = T and thus T is compact. You can look at the other choices and see that (c) is a sufficient condition which is strictly weaker than (a) and (b) while on the other hand (d) and (e) are not conditions which are strictly weaker or stronger than (c) so from here you can be sure of (c) by the exam meta. In the case of actual exam preparation it's generally better to leverage the meta to cut your decision process short so you can maximize your score rather than lingering on extra details for too long.

If you're just curious then you could consider taking S = [0,1] with the standard topology and T = (0,1) also with the standard topology. To show there is a bijection between S and T take the map which sends all points to themselves except for 0, 1, 1/2, 1/3... which get sent to 1/2, 1/3, 1/4, 1/5, ... respectively. On the other hand, the open cover {(1/n,1): n is a positive integer} of (0, 1) has no finite subcover and thus (0,1) is not compact due to the Archimedean property on R.