# If a category has both an initial and terminal object, does that mean they are the same object?

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.

• No....In $\mathbf {Sets}$ the empty set is initial, and every set with a single element is terminal. Where did you read that initial and terminal objects had to coincide?
– lulu
Sep 29, 2022 at 22:15
• Or this category $\circlearrowright i\rightarrow f\circlearrowright$
– plop
Sep 29, 2022 at 22:18
• It's the usual injection of the empty subset into any set. To stress: there are, of course, a lot of terminal objects in that category. Sure they are all naturally isomorphic, but they are not unique.
– lulu
Sep 29, 2022 at 22:26
• I think you are confused. If $I$ is an initial object then, by definition, we must have a (unique) arrow from $I$ to any other object in the category. Similarly, if $T$ is terminal, we must have a unique arrow to $T$ from any other object in the category. So of course there is a unique arrow from $I$ to $T$.
– lulu
Sep 29, 2022 at 22:34
• In your "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?" For some pair of objects I and F it might be that the compositions of morphisms passing through different A give different morphisms from I to F. However, if I and F are really going to be initial and final, respectively, they are all going to be the same composition, the unique I->F. It is part of the definition of being initial and final.
– plop
Sep 29, 2022 at 22:41

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:

1. The object $$I$$ is also terminal in $$\mathcal{C}$$.
2. The object $$T$$ is also initial in $$\mathcal{C}$$.
3. The unique morphism from $$I$$ to $$T$$ is an isomorphism.
4. 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.