The arrows from the initial object in a category are monomorphisms? 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. 
 A: This is false.  I don't have a "natural" example for you, maybe someone else can come up with one.  But here is a category for which this fails, there are exactly two objects, $\{A, I\}$, and there are exactly $6$ morphisms:


*

*$f\colon A \to I$

*$g\colon A \to I$

*$a\colon I \to A$

*$\mathrm{id}_A$

*$\mathrm{id}_I = fa = ga$

*$af = ag$


Composition is given according to the rules above.  Then $I$ is initial but $a$ is not a monomorphism because $f \neq g$ but $af = ag$.
A: 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.
A: This statement is true if and only if its dual is true:

In any category, arrows to terminal objects are epimorphisms

We have an easy counter-example in $\mathbf{Set}$ to this claim: $\emptyset \to \{ \emptyset \} $
We can find a counter-example to the original statement in $\mathbf{Ring}$: $\mathbb{Z}$ is initial, but the maps $\mathbb{Z} \to \mathbb{Z} / p \mathbb{Z}$ are not monomorphisms.

What we do have, however, is that arrows from a terminal object are monomorphisms, and similarly arrows to an initial object are epimorphisms.
A: Let $\mathcal{C}$ be a category whose objects are $\mathrm{ob}(\mathcal{C})=\{x,y\}$ and whose morphisms are
$$\begin{align}
\mathrm{Mor}_{\mathcal{C}}(x,x)&=\{\mathrm{id}_x\} & \mathrm{Mor}_{\mathcal{C}}(x,y)&=\{f\}\\\\
\mathrm{Mor}_{\mathcal{C}}(y,x)&=\{g,h\} & \mathrm{Mor}_{\mathcal{C}}(y,y)&=\{\mathrm{id}_y,k\}
\end{align}$$
where $g\circ f=h\circ f=\mathrm{id}_x$ and $f\circ g=f\circ h=k$, $\;k\circ k=k$, $\;g\circ k=g$, $\; h\circ k=h$.
Then $x$ is an initial object of $\mathcal{C}$ and $f\circ g=f\circ h$, but $g\neq h$.
(As Jim points out in a comment below his answer, the category I had used was not actually a counterexample; we need to include a non-identity morphism from $y$ to itself.)

Though you've already accepted my answer, I had been in the process of making a diagram of this category, and I might as well include it at this point:
                                       

\documentclass{standalone}
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
x \ar[bend left=40]{r}{f}
\ar[loop left,out=220,in=140,distance=1cm]{}{\mathrm{id}_x}
& y \ar[bend left=40]{l}[swap]{g} \ar[bend left=70]{l}[pos=0.47]{h}
\ar[loop right,out=322,in=38,distance=1cm]{}{\mathrm{id}_y}
\ar[loop right,out=295,in=65,distance=2.5cm]{}[swap]{k}
\end{tikzcd}
\end{document}

