>E(f) does not have to be connected even when $\ X\ $ is.

**Example:** Consider $\ S^1 := \{z\in\mathbb C : |z| = 1\}\ $ -- the unit circle; and also $\ f:S^1\rightarrow S^1\ $ such that:

$$\forall_{z\in S^1}\ f(z):= z^2$$

Then $\ E(f) = \{(u\ v)\in S^1\times S^1 : u^2=v^2\}\ $ is not conected.