0
$\begingroup$

Let $f: A\to B; \ g,h:B\to A$ and $f\circ g = I_B$ and $h \circ f = I_A$

I want to simply state that for any function $f$ if $f \circ h = I_A$ then it must be that $h = f^{-1}$ but that seems incomplete to me. What can I do for fixing this?

  • 2
    Suppose for $x,y \in A$ we have $f(x)=f(y)$. Then $x = h\circ f (x) = h\circ f(y) = y$, so $f$ is injective. That alone doesn't prove $h = f^{-1}$, unless $f$ is also surjective (which you get from $f\circ g = I_B)$. – 2017-02-27
  • 0
    If $h\circ f=1_A$ then $f$ is injective (as shown by @QiyuWen). This identity on its own does not allow you to draw conclusions like $f=h^{-1}$ or $h=f^{-1}$. – 2017-02-27
  • 0
    Can I say for the surjection that $f\circ g$ is a surjection, and because for a composition of surjections you just need the first function to be surjective, then f is surjective? – 2017-02-27
  • 0
    No. For the surjection just observe that for every $b\in B$ we can find some element in $A$ that is sent by $f$ to $b$. Note that element $g(b)\in A$ does the job: $f(g(b))=1_B(b)=b$. – 2017-02-27
  • 0
    But is this rigurous enough? I was trying to prove that if $f \circ g$ is surjective, then $f$ must be surjective. Because I understand what you say, it is basically just the definition of surjection, but how can I be rigurous enough for this? – 2017-02-27
  • 0
    I withdraw the word "No" in my former comment. You are correct in stating that surjectivity of $f\circ g$ implies surjectivity of $f$. – 2017-02-27
  • 0
    If you prove for an arbitrary $b\in B$ that some $a\in A$ exists with $f(a)=b$ then this proof of surjectivity of $f$ is rigorous enough. If $f\circ g$ is surjective then $f(g(c))=b$ for some $c\in A$ and consequently $f(b)=a$ for $b:=g(c)\in B$. That is a rigorous proof. If $f\circ g=1_B$ then of course $f\circ g$ is surjective. The proof I gave is a bit more direct, but indeed the surjectivity of $1_B$ is enough allready. – 2017-02-27

2 Answers 2

0

You just need that $h\circ f=I_A$ for some $h\colon B\to A$.

Indeed, if $f(x)=f(y)$, for some $x,y\in A$, then also $h(f(x))=h(f(y))$ and therefore $x=y$, due to $h\circ f=I_A$.

If there exists also $g\colon B\to A$ with $f\circ g=I_B$, then $f$ is surjective and $g=h$. Only in this case you can state that $g=h=f^{-1}$.

  • 0
    Yes but for surjection do I have to prove that if a composition of 2 functions is surjective, then $f$ is surjective in $f\circ g$? Or can I just declare that the identity function is surjective? – 2017-02-27
  • 0
    @TheBosco Suppose $f\circ g=I_B$ and take $b\in B$; then $b=f(g(b))$, so $b$ belongs to the image of $f$. – 2017-02-27
0

Post compose $I_B$ with $h$, then $$h\circ f\circ g = h\circ I_B=h,$$

but $h\circ f=I_A$, so $h=g$.

Since $f$ has right and left inverse, it is a bijection, in particular it is injective.