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I have a function two images $f:X \rightarrow Y$ and $g:Y \rightarrow X$

The task is to prove or disprove the following statement:

Task: If $f$ is injective and $f \circ g $ is injective, then $g$ is injective.

So I guess, that this is correct. Question: Is that correct?

$x, x' \in X, x \neq x'$ then $f(x) \neq f(x')$

and because $f \circ g$ is injective, it is true that $g(f(x)) \neq g(f(x'))$

$\rightarrow$ g is injective

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    There is something weird in your proof : generally, composition of function is defined by $(f\circ g)(x)=f(g(x))$, not $g(f(x))$... – 2017-02-18
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    In fact, the claim "if the map $x\mapsto f(x)$ is injective and the map $x\mapsto g(f(x))$ is injective, then the map $x\mapsto g(x)$ is injective" is not true in general. – 2017-02-18
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    wait - do you have an example for that – 2017-02-18

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Not exactly, but you are on the right path.

(You made one mistake: $f\circ g=f(g)$ and not $g(f)$)

You want to show that $g$ is injective, so let's take $x,y$ such that

$$g(x)=g(y).$$

Then

$$f(g(x))=f(g(y))$$

so since $f\circ g$ is injective, you have

$$x=y.$$

Remark: you don't need $f$ injective.

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    doesn't $g(x)=g(y)$ mean surjective? – 2017-02-18
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    @jublikon No, surjective means that for all $y$ there exists $x$ such that $g(x)=y$. Injective is: $g(x)=g(y)\implies x=y$, or (it is equivalent): $x\ne y\implies g(x)\ne g(y)$. – 2017-02-18