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Attempt: I need to basically show that: $$D(f)=R(g)\\ D(g)=R(f)$$

We can clearly infer the following from the information given above: $$\begin{align} D(f)=R(g\circ f) \subseteq R(g) \tag{1}\\ D(g)=R(f\circ g) \subseteq R(f) \tag{2} \end{align}$$

And we know trivially from the definition of function composition, $$\begin{align} R(f) \subseteq D(g) \tag{3}\\ R(g) \subseteq D(f) \tag{4} \end{align}$$

From $(1),(2),(3)$ and $(4)$, we get out intended result. Is this proof correct? I'd like to know if there is any other way of proving this.

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It is correct, but I would do it as follows: $D(g)\subset R(f)$ because, if $x\in D(g)$, then $x=f\bigl(g(x)\bigr)\in R(f)$ and for the same reason, $D(f)\subset R(g)$. For the rest, I would have done it as you did. But it's really just a matter of taste.

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    $\begingroup$ Always a learning experience seeing variety. $\endgroup$ – Salman Qureshi Mar 10 '18 at 13:38

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