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Let $f_A:\emptyset\to A$ be the empty function with range $A$. The definition of a bijection as applied to this function is:

$$\forall x,y \in \emptyset (x=y \implies f_A(x)=f_A(y))$$

negating you get:

$$\exists x,y \in \emptyset (x = y\land f_A(x) \neq f_A(y))$$

Which is obviously a false statement since there are no elements in $\emptyset$ at all.

I got troubled by this question when considering the empty set as an inital object of the category Set and the following theorem:

"if I is an initial object then any object isomorphic to I is also an initial object."

but since every empty function is a bijection and thus an isomorphism it follows that all the objects in Set are initial which is obviously false.

What did i miss?

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Perhaps asking questions about the empty function, a rather curious and weird beast already on its own, may be a little too much, but I'd say it is "emptily" injective. –  DonAntonio Nov 11 '13 at 13:11
    
if the domain of $f$ is the empty set then the image of $f$ is also the empty set, so the range $A$ is in fact the empty set. –  Beni Bogosel Nov 11 '13 at 13:13
    
When you don't know if $\emptyset \to A$ is surjective for all $A$, it's too early to learn category theory ... ... –  Martin Brandenburg Nov 11 '13 at 16:19
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Luckily i can spot a subjective statement when i see one. –  Saal Hardali Nov 11 '13 at 16:32
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1 Answer 1

up vote 5 down vote accepted

Your symbolic definition of bijectivity is incorrect. The condition you wrote holds because $f_A$ is a function (vacuously in the case of the empty set as domain). A function $f:A\rightarrow B$ is injective (resp. surjective) if and only if $f(x)=f(y)$ implies $x=y$ (resp. for each $z\in B$ there exists $x\in A$ with $f(x)=a$). The unique function from the empty set to any other set is injective, but can be surjective if and only if the target is empty as well.

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Oh i see now, it's just a stupid typo that eluded me. Thanks for getting me out of the confusion. –  Saal Hardali Nov 11 '13 at 13:16
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