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1. \(f\) is not surjective. There are elements in the codomain which are not in the range. For example, no \(n \in \Z\) gets mapped to the number 1 (the rule would say that \(\frac{1}{3}\) would be sent to 1, but \(\frac{1}{3}\) is not in the domain). In fact, the range of the function is \(3\Z\) (the integer multiples of 3), which is not equal to \(\Z\text{.}\)

2. \(g\) is not surjective. There is no \(x \in \{1,2,3\}\) (the domain) for which \(g(x) = b\text{,}\) so \(b\text{,}\) which is in the codomain, is not in the range. Notice that there is an element from the codomain “missing” from the bottom row of the matrix.

3. \(h\) is surjective. Every element of the codomain is also in the range. Nothing in the codomain is missed.