Trivial Group is Initial Object
Theorem
Let $\mathbf{Grp}$ be the category of groups.
Let $1 = \set e$ be the trivial group.
Then $1$ is an initial object of $\mathbf{Grp}$.
Proof
Let $\struct {G, \circ}$ be a group with identity $e_G$.
By Group Homomorphism Preserves Identity, any hypothetical group homomorphism $\phi: 1 \to G$ must satisfy:
- $\map \phi e = e_G$
Let us define the mapping $\phi$ in this way.
By Equality of Mappings, only one such mapping $1 \to G$ can exist, establishing uniqueness.
Now to verify that $\phi$ is actually a group homomorphism.
Since $1$ has only the element $e$, this is verified by:
\(\ds \map \phi e \circ \map \phi e\) | \(=\) | \(\ds e_G \circ e_G\) | Definition of $\phi$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e_G\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi e\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {e * e}\) |
where we used $*$ to denote the group operation on $1$.
Thus $\phi$ is a group homomorphism.
We have thus established that there is a unique morphism $1 \to G$ in $\mathbf{Grp}$ for all $G$.
That is, $1$ is an initial object in $\mathbf{Grp}$.
$\blacksquare$
Sources
- 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous) ... (next): $\S 2.2$: Example $2.11$: $3$