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Statement: $g:A \to B$ and $f:B \to C$ are injective functions. Then $f \circ g:A \to C$ is injective.

Attempt of proof:

Suppose by contradiction that $f \circ g$ is not injective. Then there exist $x$ and $y$, $ x \neq y$, such that $f \circ g(x) = f\circ g(y)$. Then, by definition of composition, $f(g(x)) = f(g(y))$. Since $g$ is injective, $f(x) = f(y)$. Since $f$ is injective, $x = y$. But that is a contradiction, since $x \neq y$.

I'm not sure if every step I'm doing here is alright, specialy that if $f \circ g$ isn't injective then there exist $x$ and $y$, $x \neq y$. Why there must be two points in $A$?

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    $\begingroup$ Well done (every step)! If there is only one point in $A$ then any function that has $A$ as domain is automatically injective. $\endgroup$ – drhab Apr 28 '18 at 7:20
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    $\begingroup$ Why "artificially" use proof by contradiction? Injective is defined as "f(x)=f(y)\to x=y". And we straightforwardly get from $f(g(x))=f(g(y))$ that $g(x)=g(y)$ because $f$ is injective, and then $x=y$ because $g$ is injective. $\endgroup$ – Hagen von Eitzen Apr 28 '18 at 7:27
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    $\begingroup$ Because I'm proving by contradiction the theorems I already proved directly to upgrade my skills. $\endgroup$ – user556899 Apr 28 '18 at 7:30
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If there were only one point in $A$, then $g$ is the constant function in the sense that $g$ can take value at $a\in A=\{a\}$.

Because the injectivity is defined like $\forall x,y\in D(f(x)=f(y)\rightarrow x=y)$. To get the contra, then it is $\exists x,y\in D(f(x)=f(y)\wedge x\ne y)$.

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