If I consider the function function $f:A \rightarrow B $ for $A=[{x:1\leq x \leq 2}]$ and $B= [x:x\in \mathbb{R}]$ where, $f(x)=x$. Then we can see that $f$ is injective, but it is not surjective (as all members of $B$ are not covered).

My first question is, is my reasoning valid?

If so then is it also valid that this function has a left inverse $g:B\rightarrow A$ where $g(x)=x$,thus $g(f(a))=a$, $\forall a\in A$ , which is true. But it is ture that $f$ does not have a right inverse?

  • 2
    $\begingroup$ The $g:B\to A$ is not well-defined, sincerity is not true that $g(x)\in A$ for all $x\in B.$ $\endgroup$ Sep 21 '19 at 21:16
  • $\begingroup$ You can define $g:B\to A$ which is a left inverse, but your definition is not one. $\endgroup$ Sep 21 '19 at 21:18
  • 1
    $\begingroup$ And you are correct that $f$ has no right-inverse $\endgroup$ Sep 21 '19 at 21:19
  • $\begingroup$ I understand that $g$ is not well-defined but does $f$ have a left inverse? If so then what could that function be ? $\endgroup$ Sep 21 '19 at 21:21

Any injective function $f: A \to B$ has a left inverse $g: B \to A$: for every point $b$ in $B$ pick its unique (by injectivity) $f$-preimage as $g(b)$, otherwise define $g(b) \in A$ how you like (map everything to a point e.g.). That way, $g(f(a)) = a$ for all $a$ by construction.

If $f: A \to B$ has a left inverse $g: B \to A$ then it is injective: $f(a) = f(a') \to a= g(f(a)) = g(f(a')) = a'$.

$f$ has a right inverse $g: B \to A$ iff it is surjective: if $f$ is surjective "define" $g(b)$ by by picking any element from $\{a: f(a)=b\}$ which is a non-empty set (well-order $A$ and pick the minimal element from that set, e.g.) and then $f(g(b))=b$ by construction. And given a right inverse $g$ and $b \in B$ then $f(g(b))=b$ shows that every $b$ has a pre-image $g(b)$ so $f$ is onto.

So your example cannot have a right inverse, but does have a left inverse. See monomorphisms and epimorphisms in category theory.

  • $\begingroup$ +1 Nice answer. Also nice reference there to category theory ^^. $\endgroup$
    – hal4math
    Sep 21 '19 at 22:26

So you have $f : [1,2] \to \mathbb{R}$ with $x \mapsto x$ (e.g. $f = id_{[1,2]}$). You want to find a function $g : \mathbb{R} \to [1,2]$ so that $g\circ f : [1,2] \to [1,2]$ and $g \circ f (x) = id_{[1,2]}(x) = x$. So you are right by taking $g\big|_{[1,2]}(x) = x$. Now think about what needs to happen with $x \in [1,2]^c$? Is there anything you want it to be? Yes, it must be send to $[1,2]$, but expect of that? So really it is up to what value you assign it, since all that $g$ will be subjected to in the composition $g\circ f$ is the image of $f$ and that is $[1,2]$.


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