If a category C
has initial object I
and terminal object T
, then from what I read that means that I
= T
. From the definitions of initial and terminal objects, every object A
has unique morphisms I -> A
, and A -> T
. But unless I
and T
are the same (zero) object, then won't I
have a distinct morphism from I -> A -> T
for every object A
through composition?
I haven't found this mentioned anywhere so far so I just wanted to make sure I am reading the definitions correctly.
1 Answer
If a category $\mathcal{C}$ has initial object $I$ and terminal object $T$, then from what I read that means that $I = T$.
No, initial and terminal objects need not be the same. Consider the following examples:
Consider $\mathbf{Set}$. There exists a unique initial object in $\mathbf{Set}$, namely the empty set. The terminal objects in $\mathbf{Set}$ are precisely the singleton sets (i.e., those sets that contain precisely one element). But the empty set is not a singleton set.
Consider the category $\mathbf{Ring}$ of rings. The initial object of $\mathbf{Ring}$ is $ℤ$, and the terminal object is the zero ring. These two rings are very much not the same.
Let $P$ be a partially ordered set, considered as a category in the usual way.¹ An initial object of $P$ is the same as a minimum of $P$, and a terminal object of $P$ is the same as a maximum of $P$. Unless $P$ consists of only a single element, these two will not be the same.
We have in the given arbitrary category $\mathcal{C}$ a unique morphism from $I$ to $T$. (This hold because $I$ is initial, and it also holds because $T$ is terminal.) However, the reverse is not necessarily true: there does not need to exist a morphism from $T$ to $I$. We actually have the following statement.
Proposition. Let $\mathcal{C}$ be a category with initial object $I$ and terminal object $T$. The following conditions on $I$ and $T$ are equivalent:
- The object $I$ is also terminal in $\mathcal{C}$.
- The object $T$ is also initial in $\mathcal{C}$.
- The unique morphism from $I$ to $T$ is an isomorphism.
- There exists a morphism from $T$ to $I$.
If one (and thus all) of these equivalent conditions are satisfied, then one calls the initial object of $\mathcal{C}$ (which is then also its terminal object) a zero object. Some examples of this:
- In $\mathbf{Grp}$, the trivial group is a zero object.
- Given a ring $R$, the zero module is a zero object in $R\text{-}\mathbf{Mod}$.
From the definitions of initial and terminal objects, every object $A$ has unique morphisms $I \to A$, and $A \to T$. But unless $I$ and $T$ are the same (zero) object, then won't I have a distinct morphism from $I \to A \to T$ for every object $A$ through composition?
No, the morphisms $I \to A \to T$ don’t need to be distinct.
Let us denote the unique morphism from $I$ to $A$ by $f_A$ and the unique morphism from $A$ to $T$ by $g_A$. The composite $g_A ∘ f_A$ is a morphism from $I$ to $T$, and there is only one such morphism. This means that the composite $g_A ∘ f_A$ must be the same for every object $A$.
Let us look at an explicit example:
- For every ring $R$, the unique homomorphism of rings $f_R$ from $ℤ$ to $R$ is given by $f_R(n) = n ⋅ 1_R$ for every integer $n$. The unique homomorphism of rings $g_R$ from $R$ to $0$ (the zero ring) is the constant map $r \mapsto 0$. The composite $g_R ∘ f_R$ is the map $$ ℤ \to 0 \,, \quad n \mapsto 0 \,. $$ It doesn’t matter which ring $R$ we choose, the resulting homomorphism $ℤ \to R \to 0$ is always the same.
¹ Given a partially ordered set $P$, the corresponding category $\mathcal{P}$ is defined as follows: The objects of $\mathcal{P}$ are the elements of $P$. For every two elements $x$ and $y$ of $P$, there exists at most one morphism from $x$ to $y$ in $\mathcal{P}$, and this morphism exists if and only if $x ≤ y$. The composition of morphisms in $\mathcal{P}$ is defined in the only possible way.
I
andT
are the same (zero) object, then won't I have a distinct morphism fromI -> A -> T
for every objectA
through composition?" For some pair of objectsI
andF
it might be that the compositions of morphisms passing through differentA
give different morphisms fromI
toF
. However, ifI
andF
are really going to be initial and final, respectively, they are all going to be the same composition, the uniqueI->F
. It is part of the definition of being initial and final. $\endgroup$