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So I have next theorem

Suppose that we have two functions $G: B\to C$ and $F: A\to B$ and that $G \circ F$ is injective.

Prove that $F$ is injective and $G$ has no relevance in this.

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This is simply a matter of writing out the definitions.

Suppose $f$ were not injective. Then there would be $f(x) = f(y)$ with $x \not = y$. But then $gf(x) = gf(y)$, a contradiction to the injectivity of $gf$.

$g$ has no relevance: I'd just give an example where $g$ is not injective, and an example where $g$ is injective.

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    Hmm, can you give me an exemple, in my mind thats kind of impossible. If F is injective you basically get a good smooth domain. So if G is not injective i can't get it to G o F to be injective ... – 2017-01-18
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    Let $g: \{0,1\} \to \{0\}$ by $n \mapsto 0$, and $f: \{0\} \to \{0\}$ the identity. Then $g$ is not injective, but $gf$ is the identity on $\{0\}$. – 2017-01-18