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
$\endgroup$ 42 Answers
$\begingroup$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.
Note that this is not a proof that what you wrote is wrong, but it shows you the correct version.
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.
$\endgroup$ $\begingroup$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
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