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We have the sets $A$ (set of natural number from $1$ to $4$), $B$ (set of integers from $-4$ to $0$) and $C$ (set of rational numbers between $5$ and $6$ included $5$ and $6$).

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I want to choose a set $D$ and a relation for the following:

f is a surjective map from B to A

g is an injective map from A to C

h is a bijective map from D to B

k is a relation but not a map from C to B

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Could you give me a hint how we could find such relations/maps?

Let's consider the first one. Do we have to consider a function that shifts the interval from $1$ to $4$ into the interval from $-4$ to $0$?

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Bear in mind $|A| = 4$ and $|B| = 5$ and $|C|=|\mathbb N|$.

$f: B\to A$ means every element of $A$ gets mapped to. Just do it. Any such map will do $-4\mapsto 1;-3\mapsto 3;-2\mapsto 2; -1\mapsto 4; 0\mapsto 3$ is as good as any.

$g:A\to C$ is an injective map from a set of four elements into an infinite set. It is injective which means each of the four elements of $A$ get mapped to a different element of $C$.

$h: D\to B$ is bijective. So $|D| = |B| =5$. So any set with five elements will do. Might I suggest $\{tantor, babar, pinkhonkhonk, jumbo, hathi\}$?

$k$ is a relationship but not a map from $C$ to $B$. So $k \subset C\times B$. Any will do.

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Do we have to consider a function that shifts the interval from 1 to 4 into the interval from −4 to 0?

No. You don't have any intervals. You just have a set of numbers.

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  • $\begingroup$ Ah ok!! How can we define a possible relationship $k$ ? Also the maps g and h are invertinle but the inverse of g is not a map because not all elements of the domain are mapped somewhere. Is this correct? Can you give me an example of a function whose inverse is not function? An example of a function whose inverse is a function is $f(x)=x$ or not? $\endgroup$ – Mary Star Jun 15 '18 at 3:57
  • $\begingroup$ $k=R$ so that $5\frac 12 R -1$ and $5\frac 12 R - 3$ is a relation between $C$ and $B$ that is relation that is not a function. The inverse of $f$ is not a function either because one element of $A$ is mapped from two points. $f(b) = f(\beta) = a$ so $f^{-1}(a) = \{b, \beta\}$. Not a function. $\endgroup$ – fleablood Jun 15 '18 at 5:50

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