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

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.

Answer 2

For this $f$

$$f(x) = \begin{cases} \frac 1x & \text{if } x\in (0,1] \\ 0 & \text{if }x = 0 ,\end{cases}$$

you will have $T = [1, \infty) \cup \{0\}$. (And you are right, (c) is the weakest condition along them)

Answer 3

Let $S=[0,1]$ and let $T=[0,1)\cup\{2\}.$ Can you find a bijection from $S$ to $T$?