I know that in order for a function to be invertible, it must be bijective, but does that mean that all bijective functions are invertible?
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A function is invertible if and only if it is injective (one-to-one, or "passes the horizontal line test" in the parlance of precalculus classes). A bijective function is both injective and surjective, thus it is (at the very least) injective. Hence every bijection is invertible.
As pointed out by M. Winter, the converse is not true. For example, the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x + \lfloor x \rfloor$ is injective (and therefore invertible), but not surjective.
Xander Henderson
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Yes. For every bijective function $f$ there is a unique function $f^{-1}$ so that $f(f^{-1}(x))=f^{-1}(f(x))=x$, or in other words: $f\circ f^{-1}=f^{-1}\circ f=\mathrm{id}$. Note that this is not one hundred percent precise. For this we have to state the domain and codomain of all involved functions and have to be careful to state which set $x$ belongs to.
However, your initial statement is wrong. There are invertible functions that are not bijective, e.g. $\exp:\Bbb R\to\Bbb R$, which fails to be surjective (hence not bijective).
M. Winter
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to be clear you're saying that exp is not because it is never negative? – Joseph Garvin Jun 19 '18 at 23:10
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@JosephGarvin yes, unless you state it as a function $\exp:\Bbb R\to\Bbb R^+$. Then it is bijective. – M. Winter Jun 20 '18 at 05:52
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$exp:\mathbb{R} \to \mathbb{R}$ is not invertible. But $$\text{exp}: \mathbb{R}\to \mathbb{R}^+$$ is. – Chris Christopherson Apr 17 '23 at 22:36
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@Chris This is addressed in the comments and also might depend on your definition of "invertible" and whether you use it synonymously with "bijective". – M. Winter Apr 18 '23 at 13:42
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I don't think "invertible" is a formal term with a precise definition.
Most texts take it to by synonymous with "injective" so: bijective = injective & surjective $\implies$ injective = invertible. But invertible = injective $\not \implies$ injective & surjective = bijective.
But some texts take it to be synonymous with "bijective".
The main idea is that for $f:A\to B$ to be "invertible" then for any $y \in B$ the set of all $\{x_i\} \subset A$ such that $f(x_i) = y$ should be "invertible" that is there should only be one such, not multiple, $x$ so that $f(x) = y$.
[ The set $\{x_i| x\in A, f(x_i) = y\}$ is denoted $f^{-1}(y)$. In general this is a multivalued set. But it could be an emptyset. Or a single valued set. If it's a single valued set for all $y \in {f(x)|x \in A} = f(A)$, then we can say that $f^{-1}(x)$ is a well-defined function $f^{-1}:f(A) \rightarrow A$.]
That is the very definition of "$f$ is injective". So no matter how we define "invertible" is must imply that $f$ is injective.
On the other hand it's ambiguous whether it's possible that there are no $x\in X$ so that $f(x) = y$. (In other words $y \in B$ but $y \not \in f(A)$; or in other words $f^{1}(y) =\emptyset$.) It could be argued it's not invertible if $f^{-1}(x) = \emptyset$ rather than a single element. Or it could be argued that it doesn't matter and we are only interested in $y \in f(A)$.
At any rate "invertible means that for all $y \in f(A)$ then $f^{-1}(y)$ is a single element; not a multiple valued set. In other words "invertible" $\implies$ "injective".
fleablood
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Oh I'm so late to see and upvote your answer... I just asked a question about this issue... – linear_combinatori_probabi Apr 10 '18 at 15:30
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And there is a typo, which you meant $f^{-1}(y)=\emptyset$ right? Anyway, I'm very happy to see people care what I care. In case you want to know the link of my question – linear_combinatori_probabi Apr 10 '18 at 15:35