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This problem is in the book I am studying for proofs. A theorem stated in the book is that a function $f:A \rightarrow B$ is bijective if it is injective/surjective. Also the Domain of $f$ has to equal $A$.

While proving that a composition of functions is bijective if the functions themselves are bijective, the author proves in the end that given functions $f:A\rightarrow B$ and $g:B \rightarrow A$ the domain of $g \circ f=A$ by using the following logic:

$Dom(f)=A$ and $Dom(g)=B$.

Let $a \in A$. Then $a \in Dom(f)=A$ so $\exists b \in B$ such that $f(a)=b$.

But $b \in B=Dom(g)$ so $\exists c \in C$ such that $g(b)=c$ then

$(g \circ f)(a)=g(f(a))=g(b)=c$ So $a \in Dom(g \circ f)$

Im sorry if this is obvious but how does this prove the domain of $g \circ f=A$ ?

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  • $\begingroup$ I'm not sure to understand the question. Writing $f \colon A \to B$ means that $f$ is a function whose domains is $A$. So, the fact that domain of $g \circ f$ is $A$ has nothing to do with the fact that $f$ and $g$ are bijective. Instead, it has to do with the definition of composition of two (bijective or not) functions $f \colon A \to B$ and $g \colon B' \to C$, where $B \subseteq B'$. $\endgroup$ Commented Jul 25, 2019 at 7:39
  • $\begingroup$ @Taroccoesbrocco the book says that to prove a function $f:A \rightarrow B$ is bijective I need to prove the $domain(f)=A$ this was part of a proof in the book that a composition of functions $g \circ f$ is bijective. So one part of the proof was prove $Dom(g \circ f)=A$ $\endgroup$ Commented Jul 26, 2019 at 2:45

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In order to show that domain of $gof$ is $A$ we need to show that for every $a\in A $ , $gof(a)$ is well defined.

For an arbitrary element $a$ of $A$, f(a) is a well defined element of $B$ and $g(f(a)) $ is a well defined element of $C$.

Thus (gof)(a) = g(f(a)) is well-defined.

That makes $A$ to be the domain of $fog$

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