prove that composition $g$ of $f$ is bijective then $f$ is injective and $g$ is surjective


Your proof for $f$ being injective and $g$ being surjective are correct. For the counterexample, let $A = C = \{0\}$, $B = \{0,1\}$, and define $f$ such that $f(0) = 0$, and $g(0) = g(1) = 0$. Then clearly $f$ is not surjective and $g$ is not injective, but $g \circ f : \{0\} \to \{0\}$ with $g \circ f(0) = 0$ is clearly bijective.


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K. Gibson
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K. Gibson

Updated on January 19, 2020


  • K. Gibson
    K. Gibson almost 4 years

    Claim: If $g \circ f: A \to C$ is bijective then where $f:A \to B$ and $g:B \to C$ are functions then $f$ is injective and g is surjective. Also give an example where $g \circ f$ is bijective but $f$ is not surjective and $g$ is not injective.

    Proof attempt:

    Well if $g \circ f$ is bijective then $g \circ f$ is injective. Therefore $f$ is injective:

    Suppose $$f(x)=f(y)$$

    $$g(f(x))=g(f(y))$$ which means

    $$(g \circ f)(x) = (g \circ f)(y)$$ since $g \circ f$ is injective we have $x = y$

    Also if $g \circ f$ is bijective then $g \circ f$ is surjective and therefore $g$ is surjective

    Suppose that $$g \circ f$$ is surjective take any $$y \in C$$

    Since $$g \circ f$$ is surjective there exists an $$a \in A$$

    such that $$(g \circ f)(a) =y$$ so

    $$g (f(a)) = y$$ set

    $$b=f(a) \in B$$ Then $$g(b)=g(f(a))=y$$

    Therefore g is surjective.

    firstly just to verify, my proofs answer the question right? Secondly is there a more direct and faster method?

    Now for the second part:

    $A = \{1\}, B=\{2,3\}, C\{4,5\}$

    Define $f: A \to B$ as $f\{1\} \to \{2\}$ which is clearly not surjective because 3 is not mapped. Define $G\{2,3\} \to \{4,4\}$ Thus $g$ is not injective however my problem is that I don't think $f \circ g$ is bijective I just cannot figure out how to rectify a composition where one function is not surjective and the other is not injecitive and get a bijection