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If $A = \{1,2,3,4\}$ and $B = \{6,7\}$ and following are relations from $A$ to $B$, then State Whether these are functions or not?

If these are functions what kind of functions are they $$ \begin{split} R_1 &= \{(1,5),(2,7),(3,6)\}\\ R_2 &= \{(1,6),(2,6),(3,7),(4,7)\}\\ R_3 &=\{(1,6),(2,6),(3,6),(4,6)\} \end{split} $$

so can anyone explain what are those first $A$ and $B$ functions and then $R_1$ to $R_3$, what kind of functions are they really?

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    Are you sure you have the question right? $R_1$ is not a relation from $A$ to $B$, because $5 \notin B$. – 2017-02-03

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First we need the definition of a function:

A function $f$ from $A$ to $B$ is a subset of the Cartesian product $A × B$ subject to the following condition: every element of $A$ is the first component of one and only one ordered pair in the subset. (Source: Wikipedia)

Let us use $R_1 = \{(1,5),(2,7),(3,6)\}$ as an example. Here, the first component of each pair does indeed belong to $A$ and no element if $A$ is repeated. However, the pair $(1,5)$ tells us this is not a function because $5 \not\in B$.

Now to figure out what type of function it is. I believe it is asking whether the function is injective (one-to-one) or surjective(onto) or neither. If it is both injective and surjective, it is called a bijective function. Injective means that there can not be two elements $x_1, x_2 \in A$ that map to the same element $y \in B$. That is, your relation can not have two ordered pairs that have different first values but same second values. Let us assume $5\in B$ so that we can examine what type of function this is. So now that we have $B = \{5,6,7\}$, we see that $R_1$ is injective since it maps unique elements of $x$ to different elements in $B$. Now to check if it is surjective. Surjective means that every element in $B$ is mapped to by atleast one element in $A$. Another way to say this is that if $y\in B$ then there exists an ordered pair $(x,y)$ for some $x\in A$.Since all the elements in $B$ are mapped to, $R_1$ is also surjective. This implies that $R_1$ is a bijective function. (Note; I changed the codomain $B$ just so that I can explain to you different types of functions. Please change the set $B$ back to $\{6,7\}$ and try it out yourself.)

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    So does it means all element of A and B must be present in this function to be called function? – 2017-02-03
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    and what type of function is it? – 2017-02-03
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    No. The definition of a functions says that it is a subset of the Cartesian product $A \times B$. Since $5 \not\in B$, there can not exist an ordered pair in $A \times B$ such that $5$ is the second element in the ordered pair. – 2017-02-03
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    I believe it is asking if the function is injective (one-to-one) or surjective (onto). If it is both injective and surjective, it is called a bijective function. – 2017-02-03
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    So R1 is injective function and R2 and R3 is surjective function? – 2017-02-04