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Suppose that $f: A \to B$ is injective, $A = \{5, 7\}, B = \{9, 10\}$ and $f(5) = 9$, $f(7) = 10$.

Now construct $g: B \to A | g(9) = 7, g(10) = 5$.

Does this function $f$ have a left inverse $g$, i.e. $g(f(a)) = a$ for all $a \in A$?

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    $\begingroup$ How can $f(5) = 9$ be if $9 \notin B$? Or $f(7) = 10$ if $7 \notin A$? $\endgroup$ – Martin R Feb 10 at 17:47
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    $\begingroup$ You have confused notation here. In your definition $f$ is not $A \rightarrow B$. $\endgroup$ – Alvis Nordkovich Feb 10 at 17:48
  • $\begingroup$ @MartinR please see edit. $\endgroup$ – Jossie Calderon Feb 10 at 18:03
  • $\begingroup$ @Alvis see edit. $\endgroup$ – Jossie Calderon Feb 10 at 18:03
  • $\begingroup$ $g(f(5)) = g(9) = 7 \ne 5$ and $g(f(7)) = g(10) = 5 \ne 7$ $\endgroup$ – lightxbulb Feb 10 at 18:06
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The function $g(9) = 7, g(10) = 5$ is not an inverse to $f$, as you can easily verify by computing $g(f(5)) = g(9) = 7 \ne 5$. On the other hand both $g$ and $f$ are bijections so they have well-defined inverses: $$f^{-1}(9) = 5, f^{-1}(10) = 7$$ $$g^{-1}(7) = 9, g^{-1}(5) = 10$$ If you want to construct an inverse to a discrete injective function $h:C\rightarrow D$ then $\forall c \in C : h(c) = d \in D$ define $h^{-1}(d) = c$. Then by construction the function $h^{-1}$ satisfies $h^{-1}(d) = h^{-1}(h(c)) = c$.

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  • $\begingroup$ My error was in thinking that it was necessary for f to have a left inverse. $\endgroup$ – Jossie Calderon Feb 10 at 18:58
  • $\begingroup$ but why does f need to be one to one for it to hold true? $\endgroup$ – Jossie Calderon Feb 10 at 19:15
  • $\begingroup$ would this suffice? let f: A-> B, and a in A. Then f(a) = b in B. Let g: B-> A s.t. g(b) = f^-1(b). If f^-1(b) =/= a then two elements exist in the fiber of B, but that violates the claim that f is injective. Thus f has a left inverse. $\endgroup$ – Jossie Calderon Feb 10 at 19:21
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    $\begingroup$ @JossieCalderon Well you have an inverse precisely because of the injectivity yes. If you want an example specific to right-left inverse: math.stackexchange.com/questions/507279/… Suppose that a function is not injective. That is you have $a_1,a_2 \in A$ such that $a_1\ne a_2$ but $f(a_1) = f(a_2) = b \in B$. Then you can't have an inverse function of this form, since it will violate the definition of function. However the inverse set mapping always exists (the set $f^{-1}(b) = \{a_1,a_2\}$, however that's a set and not a function anymore). $\endgroup$ – lightxbulb Feb 10 at 19:25
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Your question caused a lot of confusion. So let us prove the following which is probably that what you really mean:

A function $f : A \to B$ is injective if and only if it has a left inverse $g : B \to A$. Here, $A,B$ are any two sets.

1) Let $f$ be injective. Choose any $a_0 \in A$ and define $$g : B \to A, g(b) = \begin{cases} a & b = f(a) \in f(A) \\ a_0 & b \in B \setminus f(A) \end{cases} $$ Note that by injectivity of $f$ for each $b \in f(A)$ there exists a unique $a \in A$ such that $b = f(a)$. By definition $(g \circ f)(a) = g(f(a)) = a$.

2) Let $g$ be a left inverse for $f$. If $f(a) = f(a')$, then $a = g(f(a)) = g(f(a')) = a'$. This means that $f$ is injective.

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