function that preserves compactness 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.
Take $S = [0, 1]$ with the usual topology and $T = [0, 1]$ with the discrete topology (all subsets are open) and take $f$ to be the identity function. $T$ is not compact: $\{\{x\} : x \in [0, 1]\}$ is an open cover with no finite subcover.
For anyone who is still reading this thread, the answer is actually A because continuity only guarantees that the image $f(S)\subseteq T$ is compact. However, we still need that image to be equal to the set $T$ to guarantee that $T$ will be compact, which we can satisfy with surjectivity, a condition only given by choice A.