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I have two simple questions about the category theory.

  1. In any category, is $Hom(A, B)$ always nonempty? In some typical categories, it seems right but the definition of morphism does not give any information.

  2. In the category $\textbf{Set}$ of all sets, what is in $Hom(\varnothing, A)$ and $Hom(A, \varnothing)$?

Please anybody help me. Thanks in advance.

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  • $\begingroup$ It can be empty, and the a relation on some empty set is empty, so there is an empty function. $\endgroup$ – user40276 Feb 2 '14 at 16:29
  • $\begingroup$ $\hom_{\mathsf{Ring}}(\mathbb{Q},\mathbb{Z})=\emptyset$. $\endgroup$ – Martin Brandenburg Feb 2 '14 at 21:19
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Hom(A,B) can be empty. For example, the category with only two objects and only identity morphisms is a perfectly valid category. For a more "natural" example, the path groupoid of a space with two different path components will have no morphisms from any object in one component to the other component. (Actually the first example is a specialization of the second to the discrete space with two points)

In Set, Hom($\emptyset,A$) consist of a single morphism for each $A$, the empty function. $\emptyset$ is the initial object in the category of sets.

Hom($A,\emptyset$) is always empty unless $A = \emptyset$, in which case it has only the identity morphism.

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    $\begingroup$ A function $f:A \to B$ is by definition a subset of $f \subset A\times B$ such that for all $a \in A$ there is a unique $b \in B$ so that $(a,b) \in f$. There can be no function from $A \to \emptyset$ if $A$ is non empty, because there are no $b \in \emptyset$ to pair with a given $a \in A$. However, there is a function from $\emptyset \to A$ because the universal quantifier is vacuously satisfied: there are no $a \in \emptyset$ to place any conditions on the relation. $\endgroup$ – Steven Gubkin Feb 2 '14 at 17:56
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    $\begingroup$ Codomains and domains aren't symmetric in this way. Every value of the empty function is in A vacuously; that's enough for it to be the codomain. But a function must map every element in its domain somewhere. $\endgroup$ – Malice Vidrine Feb 2 '14 at 17:58
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    $\begingroup$ $A\times\emptyset =\emptyset$. $\endgroup$ – Malice Vidrine Feb 2 '14 at 18:11
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    $\begingroup$ Also, in any Cartesian closed category $A\times 0\cong 0$, so it's not just a quirk of $\mathbf{Set}$ either. $\endgroup$ – Malice Vidrine Feb 2 '14 at 18:19
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    $\begingroup$ @user40276 Perhaps this will help convince you. It is true for all finite sets that if $|A|=n$ and $|B|=m$ then $|A\times B|=nm$ where $n,m\in\mathbb{N}$. It follows that $|A\times\emptyset|=0$ for all finite sets $A$ and we know that the only set with zero elements is the empty set. Hence $A\times\emptyset=\emptyset$. $\endgroup$ – Dan Rust Feb 2 '14 at 18:48

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