TITLE: The arrows from the initial object in a category are monomorphisms?
QUESTION [15 upvotes]: Let $\mathbb{A}$ be a category and $I \in \mathbb{A}$ its initial object ($\forall A \in \mathbb{A}. \exists ! f:I \longrightarrow A$). 
For $A \in \mathbb{A}$, prove that $f:I \to A$ is a monomorphism ($\forall B \in \mathbb{A}, g,h:B \longrightarrow I. f\circ g=f\circ h \implies g=h$).
I couldn't prove it nor refute it.

REPLY [5 votes]: It is often a good idea to look at examples first before trying to prove something. It already fails for the category of unital rings, here $\mathbb{Z}$ is initial and $\mathbb{Z} \to R$ is only a monomorphism when $R \neq 0$ and the characteristic of $R$ is zero. More generally, we have the following result:

Let $I$ be an initial object of a category. For the statements

$I$ is strict initial, i.e. every morphism $X \to I$ is an isomorphism.
All morphisms $X \to I$ are equal.
Every morphism $I \to Y$ is a monomorphism.

we have 1. => 2. => 3.

Proof: 1. => 2. Let $f,g : X \to I$ be morphisms. By assumption $f$ is an isomorphism, so we may look at $g f^{-1}$. This is an endomorphism of $I$, hence the identity. 2 => 3. is trivial.