# Question regarding proof that the composite mapping of two one-one maps is one-one

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)

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 $$x_0=x_1$$.
• Actually , I meant, is my proof considerable for proving that $(g \circ f)$ is one to one ? May 30, 2019 at 9:49