assume that $S$ and $T$ are sets, $\alpha:S\to T$, and $\beta:S\to T$. complete the following statements:

$\alpha$ is not onto iff there is a $y$ in $T$ such that for all $x$ in $S$, $\alpha(x)\neq y$.

$\alpha$ is not one-to-one iff there is $x,y\in S$ where $x\neq y$ such that $\alpha(x)=\alpha(y)$

$\alpha\neq\beta$ iff there is $x\in S$ such that $\alpha(x)\neq\beta(x)$.

$\beta$ is one-to-one and onto iff for each $y\in T$, there is $x\in S$ such that $\beta(x)=y$ and for all $x,y\in S$ where $x\neq y$, $\beta(x)\neq\beta(y)$.am i right

Question

assume that $S$ and $T$ are sets, $\alpha:S\to T$, and $\beta:S\to T$. complete the following statements:

$\alpha$ is not onto iff there is a $y$ in $T$ such that for all $x$ in $S$, $\alpha(x)\neq y$.

$\alpha$ is not one-to-one iff there is $x,y\in S$ where $x\neq y$ such that $\alpha(x)=\alpha(y)$

$\alpha\neq\beta$ iff there is $x\in S$ such that $\alpha(x)\neq\beta(x)$.

$\beta$ is one-to-one and onto iff for each $y\in T$, there is $x\in S$ such that $\beta(x)=y$ and for all $x,y\in S$ where $x\neq y$, $\beta(x)\neq\beta(y)$.am i right

ash2326 · Answer

Yeah, you're correct