# logicsets

• November 17th 2008, 09:56 AM
srk619
logicsets
can anyone do this question

The sets A, B and C are given by
A = {1, 2, 3, 4},
• November 17th 2008, 10:18 AM
clic-clac
A diagram is easy but with a pen :)

Let see b)

$f (1) = y, f (2) = x, f (3) = z$ and $f:A\rightarrow B=\{x,y,z\}$, so $f$ is surjective. But since $A$ has more elements than $B$, $f$ can't be injective, so can't be bijective.

$g(x) = a, g(y) = b$ and $g:B\rightarrow C=\{a,b\}$, so $f$ is surjective. But since $B$ has more elements than $C$, $g$ can't be injective, so can't be bijective. (yeah the proof is quite similar)

$h$ is the identity map on $C$, it is bijective (i.e. injective and surjective).

$k(a) = 2, k(b) = 2$ That means $k:C\rightarrow A$ is not injective. Furthermore, as $A$ has more elements than $C$, $Range(k) \neq A$ so $k$ is not surjective, in conclusion, $k$ is "none of these".