I'm studying Aluffi's Chapter 0 and think I could use some basic ideas about terminal objects.

Any morphism from a terminal object to itself is the identity. To prove this, we just observe that there can be only one morphism from an initial object $A$ to any other object, and it must have the identity map to itself. So if $\| \text{Hom}(A,B) \| = 1$ for any object $B$. In particular, $\text{Hom}(A,A)$, must contain $\text{id}_A$, so if $\phi \in \text{Hom}(A,A)$, then $\phi$ must be the identity.

For final objects the proof would be similar. In practice, if I could get a composition of functions from an initial object to itself, I could say that function was the identity, thus an isomorphism, etc.

If an object is isomorphic to a final/initial object, it itself is final/initial. To prove this, suppose $A$ is initial, $\phi: A \rightarrow B$ is an isomorphism, and $C$ is another object in our category. A morphism $A \rightarrow C$ always exists, so the composition $B \rightarrow A \rightarrow C$ where the first arrow is $\phi^{-1}$ and the second arrow is the morphism given by $A$ being initial, always exists. If $f$ were any other morphism $f:B \rightarrow C$, then the composition $A \rightarrow B \rightarrow C$ is a morphism from $A \rightarrow C$ so it must be unique.

In practice, I would use this to conclude an object is initial/terminal if I could find an isomorphism with another initial/final object.

This is just a sketch, but is it accurate? Edited for clarity and filling details.

  • 2
    $\begingroup$ It's not quite right: in the first part, the property of terminal objects is that there is only one morphism from any object to the terminal object, not that there is only one morphism from the terminal object, so you need to change your justification. In the second part, the sketch of the uniqueness is a bit sparse, but more seriously, you never prove that a morphism exists in the first place. You need to show that for any object $C$, there is a morphism $B\to C$, and that it is unique; you seem to only be proving the latter. $\endgroup$ Commented Sep 9, 2020 at 16:34
  • $\begingroup$ @ArturoMagidin I believe that the book uses “terminal” to denote either an “initial” or “final” object. Indeed, the OP does the proof for an initial object and then remarks that it's essentially the same for a final object. $\endgroup$
    – egreg
    Commented Sep 9, 2020 at 17:26
  • 1
    $\begingroup$ @egreg: Fair enough; I've never encountered that nomenclature. $\endgroup$ Commented Sep 9, 2020 at 17:30

1 Answer 1


I guess that the book uses “terminal object” to denote either an “initial” or a “final” object.

As usual, you don't need to do both proofs for initial and final objects, because an initial object in $\mathcal{C}$ is a final object in $\mathcal{C}^{\mathrm{op}}$ and conversely.

Your proof is correct, but too long. Let $A$ be an initial object and let $f\colon A\to A$ be a morphism. Since $\mathrm{id}_A$ is a morphism with the same source, namely $A$, and target, we have that $f=\mathrm{id}_A$, by uniqueness.

Suppose now that $B$ is isomorphic to an initial object $A$; let $b\colon A\to B$ be an isomorphism. Take any object $C$ and two morphisms $f,g\colon B\to C$. Then $fb,gb\colon A\to C$ are two morphisms, so they are equal. From $fb=gb$ we obtain $f=g$ by composing with $b^{-1}$. This proves uniqueness. Existence follows from the unique morphism $h\colon A\to C$, so we have $hb^{-1}\colon B\to C$.


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