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I have come across a question regarding composite mappings.

The question is as follows-

Prove that the composite mapping of two one-one maps is one-one.

My proof is as follows-

let us consider two one to one mappings $f : X \to Y$ and $g : Y \to Z$.

Since $g $ is one to one, then if

${\rm z}_0 = {\rm z}_1$

$\implies$ $g({\rm y}_0) = g({\rm y}_1)$

$\implies$ $g(f({\rm x}_0)) = g(f({\rm x}_1))$

$\implies$ $(g \circ f){\rm x}_0 = (g \circ f){\rm x}_1$

$\implies$ ${\rm x}_0 = {\rm x}_1$

Hence, $(g \circ f)$ is one to one function.

Is my proof considerable ?

Can I be provided with a more formal and a detailed proof ? ( I would like to be as pedantic as possible)

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1 Answer 1

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To prove that $f\circ g$ is one-one you have to start with the equation $f(g(x_0))=f(g(x_1))$ and you have to deduce that $x_0=x_1$, The first step is to us the fact that $f$ is one-one so we get $g(x_0)=g(x_1)$. The next step is to us the fact that $g$ is one-one so we get $x_0=x_1$.

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  • $\begingroup$ Is my proof considerable? $\endgroup$ May 30, 2019 at 8:55
  • $\begingroup$ Unfortunately, no. The starting point of the proof has to be exactly as I have stated. $\endgroup$ May 30, 2019 at 8:58
  • $\begingroup$ Actually , I meant, is my proof considerable for proving that $(g \circ f)$ is one to one ? $\endgroup$ May 30, 2019 at 9:49

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