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....


  • 1
    $\begingroup$ 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. $\endgroup$ Jun 8, 2018 at 23:41
  • 2
    $\begingroup$ 2 depends on the axiom of choice ... But more importantly, something that is true for one specific $f$ is not necessarily true for every $f$. $\endgroup$ Jun 8, 2018 at 23:43
  • 1
    $\begingroup$ You haven't actually provided counterexamples for 3 or 4. $\endgroup$
    – user169852
    Jun 9, 2018 at 0:42
  • $\begingroup$ that was my only logics @Bungo $\endgroup$
    – jasmine
    Jun 9, 2018 at 0:44

2 Answers 2


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.


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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