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Let $A$ and $B$ be sets and let $f:A\rightarrow B$ and $g: B\rightarrow A$ be functions.

  1. Suppose that $f$ is injective, and $g$ is a left inverse of $f$. Prove that $g$ is surjective.

So I have this:

Let $a \in A$, then $g∘f(a)=g(f(a))=g(b)=a$. Therefore this means that for any $a \in A$ there is a $b$ such that $g(b)=a$. Hence, $g$ is surjective.

  1. Suppose that $f$ is surjective, and that $g$ is a right inverse of $f$. Prove that g is injective.

For all $y \in B$, there is an $x \in A$ such that $f(x)=y$. Then $(f∘g)(y)=I_A(y)=y$. Let be $x=g(y)\in A$. Then $x=g(y) \in A$, $f(x)=y$ Did I prove that g is injective?

  1. Suppose that $f$ is bijective, and that $g$ is an inverse of $f$. Prove that g is bijective.

How do I prove the last one?

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  • $\begingroup$ A function $g$ is an inverse of $f$ if and only if... $\endgroup$ – user228113 May 7 '17 at 1:27
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    $\begingroup$ 1) is good but I don't like to flow g(f(a))=g(b)=a. It's acceptable but it seems a jump. And b was never introduced. I'd say: "Let $a \in A$ and let $b = f(a)$. Then $g(b) = g(f(a)) = a$. So $g$ is surjective" Its the same thing but flows more natural with the definitions and conclusion. $\endgroup$ – fleablood May 7 '17 at 2:20
  • $\begingroup$ Um... you'd do 3) exactly to same way you did 1 and 2. Except... you don't have to.. $f$ is bijective means $f$ is surjective and injective and g is a right and left inverse and you've proven that therefore g if surjective by 1) [f is injective and g is a left inverse] and that g is injective by 2) [f is surjective and g is a right inverse] so ... what's left to prove? $\endgroup$ – fleablood May 7 '17 at 2:25
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$2$. I did not understand your proof. Let's prove it. To prove that $g$ is injective, we must show $$g(x)=g(y)\Rightarrow x=y$$ for all $x,y\in B$. If there are $x,y\in B$ such that $g(x)=g(y)$, then $$x=(f\circ g)(x)=f(g(x))=f(g(y))=(f\circ g)(y)=y.$$

$3$. Since $g$ is an inverse of $f$, then $g$ is a left and right inverse of $f$. For $1$ and $2$, we have that $g$ is surjective and injective, that is, $g$ is a bijection.

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Let $A$ and. $B $ be sets and $f : A \rightarrow B$ and $g : B \rightarrow A$ be functions.

1)Suppose $f$ is injective, and $g$ is a left inverse of $f$. Prove that $g$ is surjective.

NB. Possibly $f(A) \subset B$. So, a little more precise, $g$ is left inverse of $f$ on $f(A)$.

Let $a \in A $, and $b = f(a)$, then $g(b) = g(f(a)) = a$ , since $g$ is left inverse of $f$ on $f(A)$, hence $g$ is surjective.

2) Suppose $f$ is surjective and $g$ is right inverse of $f$. Prove that $g$ is injective.

$f(A) = B$, since $f$ is surjective.

Let $x,y \in B$, $x \ne y$ , and consider $g(x), g(y)$.

Since $g$ is right inverse of $f$:

$x = f(g(x)) \ne y = f(g(y))$, hence the arguments of the function $f$ are not equal, i.e. $g(x) \ne g(y)$, $g$ is injective.

3) Answer by Rafael Hollande.

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