# which of the following is necessarily true for a function $f : X \rightarrow Y$?

which of the following is necessarily true for a function$f : X \rightarrow Y$?

1) if $f$ is injective ,then there exists $g : Y \rightarrow X$ such that $f(g(y) =y$ for all $y \in Y.$

2) if f is surjective ,then there exists $g : Y \rightarrow X$such that $f(g(y) =y$ for all $y \in Y.$

3) if $f$ is injective and $Y$ is countable then X is finite.

4) if $f$ is surjective and $X$ is uncountable then $Y$ is countably infinite

My attempts : option 3 is wrong take $f$ :N $\rightarrow$ N

option $4$ is wrong take $f : R \rightarrow R$

option $1$ is true and option $2$ is also Trues as both are True because take $f(x) = x$

Is my answer is correct or not pliz verified its....

thanks

• 1 and 2 are not both true. You can show something is false by counterexample, but you can't show something is true by example. Jun 8, 2018 at 23:41
• 2 depends on the axiom of choice ... But more importantly, something that is true for one specific $f$ is not necessarily true for every $f$. Jun 8, 2018 at 23:43
• You haven't actually provided counterexamples for 3 or 4.
– user169852
Jun 9, 2018 at 0:42
• that was my only logics @Bungo Jun 9, 2018 at 0:44

1) If $f$ is injective it means that you can get the input back given the output, which means the existence of $g$ giving or the identity by composing the other way around.
2) $f$ being surjective means that for every possible output $y$ you can choose (axiom of Choice as mentioned in the comments) a preimage and call it $g (y)$. Then indeed $f (g (y))=y$ by construction.
f is surjective ,then there exists $g : Y \rightarrow X$such that $f(g(y) =I$ for all $y \in Y.$
take X ={1} and$Y$ ={$1,2$}, as for option $1)$ $f$ is injective that mean $f(1) = 1$ but f is not onto so it is false
so option $2$ is true