yes to all. the identity function is an example for all of these

$\displaystyle f: \mathbb{R} \mapsto \mathbb{R}$ defined by $\displaystyle f(x) = x$ for all $\displaystyle x \in \mathbb{R}$ ...........it is not R "implies" R by the way, it is f:R --> R, or f maps from R to R, or f: R to R or something

i leave it to you to check that this is a reflexive, symmetric and transitive relation (aka, an equivalence relation)

do you know the definitions for one-to-one and onto?

remember (g o f)(x) = g(f(x)). what does it mean for this function to be one-to-one? what does it mean for f to be onto? playing around with these definitions should help

what definitions are you working with? i'd say use induction on the cardinality of one of the sets. but maybe the definition you are using will give us an easier way. you may also want to look at the proof of theorem 1

here