Does this injective function have an inverse?

Question

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 $a \in A$?

How can $f(5) = 9$ be if $9 \notin B$? Or $f(7) = 10$ if $7 \notin A$? — Martin R, Feb 10 '19 at 17:47
You have confused notation here. In your definition $f$ is not $A \rightarrow B$. — Alvis Nordkovich, Feb 10 '19 at 17:48
It is unclear to me what you are asking. You define a function $f : A \to B$, and it is obviously a bijection. Hence it has an inverse $f^{-1}$, but it is not the function $g$ which you have defined. — Paul Frost, Feb 10 '19 at 18:16
@PaulFrost does it have a left inverse such that g is the identity map? — Jossie Calderon, Feb 10 '19 at 18:19
@lightxbulb That is not the definition of an inverse according to Dummit and Foote. a in A such that f(a) in C (a subset of B) — Jossie Calderon, Feb 10 '19 at 18:20
@JossieCalderon Where exactly did I define an inverse? The inverse would be $f^{-1}(9) = 5, f^{-1}(10) = 7$. If you're talking about the $g$, it was just to show that the claim $g(f(a))=a$ does not hold for any $a \in A$. — lightxbulb, Feb 10 '19 at 18:25
The problem in Dummit and Foote says if f is an injection, show f has a left inverse. You just helped me prove a counterexample, so are they wrong? — Jossie Calderon, Feb 10 '19 at 18:27
@JossieCalderon They are not wrong, I just defined the inverse $f^{-1}$. It's just that the $g$ you picked is not it. — lightxbulb, Feb 10 '19 at 18:30
Perhaps your question is this: Does every injective function $f : A \to B$ (where $A, B$ are arbitrary sets) have a left inverse $g : B \to A$? — Paul Frost, Feb 10 '19 at 18:31

score 1 · Accepted Answer · answered Feb 10 '19 at 18:47

1

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 $h^{-1}(d) = h^{-1}(h(c)) = c$.

answered Feb 10 '19 at 18:47

lightxbulb

2,047

My error was in thinking that it was necessary for f to have a left inverse. – Jossie Calderon Feb 10 '19 at 18:58
but why does f need to be one to one for it to hold true? – Jossie Calderon Feb 10 '19 at 19:15
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.
– Jossie Calderon Feb 10 '19 at 19:21
1

@JossieCalderon Well you have an inverse precisely because of the injectivity yes. If you want an example specific to right-left inverse: https://math.stackexchange.com/questions/507279/example-of-left-and-right-inverse-functions 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). – lightxbulb Feb 10 '19 at 19:25

score 0 · Answer 2 · answered Feb 10 '19 at 23:30

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.

Does this injective function have an inverse?

2 Answers2