Prove that if $g \circ f$ is injective, then $f$ must also be injective.
$g(f(x))$ is injective.
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Then, $x\neq x’ \implies (g\circ f)(x) \neq (g\circ f)(x’)$.
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Suppose for the sake of contradiction: $f(x)=f(x’) \nRightarrow x=x’$
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Then this implies that it is logically possible that for some $x$ and $x’$:
$f(x)=f(x’) \implies (g\circ f)(x) \neq (g\circ f)(x’)
\implies g(f(x))\neq g(f(x’))$
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But if $f(x)=f(x’)$, then $f(x)=f(x’) \implies g(f(x))\neq g(f(x))$ which is a contradiction.
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Therefore $f(x)=f(x’) \implies x=x’$, that is, $f$ is also injective.
I’ve numbered the lines to make it easier to reference a mistake if I’ve made one. My concern is (3.). I think it should be correct to say that if (2.) holds, then that suggests that it is logically possible that $x\neq x’$ could hold under $f(x)=f(x’)$.
Best Answer
The proof is, in principle, correct, but all the use of logic symbols are wrong. Proof: Suppose that $f$ is not injective. Then for some $x\ne x'$ we have $f(x)=f(x')$. But then $g(f(x))=g(f(x'))$, so $g\circ f$ is not injective, a contradiction.