In preadditive categories, initial objects are terminal

Question

I want to prove: in a preadditive category $\mathscr{C}$, if an object of $\mathscr{C}$ is initial, thent it is also terminal (hence a zero object).

So, let's start a proof. I am reading Borceux's proof, from Handbook of Categorical Algebra, Vol.1.

Let 0 be an initial object of $\mathscr{C}.$ The set $\mathscr{C}$(0,0) has a single element, which proves that $1_{\textbf{0}}$ is the zero element of the group $\mathscr{C}$(0,0). Given an object $C$, $\mathscr{C}$(C,0) has at least one element: the zero element of that group. This is where my understanding arrives. What follows is impossible to me to be understood, so I am asking help for that.

But if $f:C\rightarrow$0 is any morphism, then $f=1_{\textbf{0}}\circ f$ must be the zero element of $\mathscr{C}$(C,0), since $1_{\textbf{0}}$ is the zero element of $\mathscr{C}$(0,0) (bilinearity of the composition.)

Answer 1

We can proceed as follows:

1. The identity morphism of $\mathbf{0}$ is also the zero element of the abelian group $\mathscr{C}(\mathbf{0}, \mathbf{0})$. This holds because $\mathscr{C}(\mathbf{0}, \mathbf{0})$ consists of only a single element, since $\mathbf{0}$ is initial in $\mathscr{C}$, and both the zero element and the identity morphism need to be this single element.

2. Composition of morphisms in $\mathscr{C}$ is bilinear. That is, the map $$ \mathscr{C}(Y, Z) × \mathscr{C}(X, Y) \longrightarrow \mathscr{C}(X, Z) \,, \quad (g, f) \longmapsto g ∘ f $$ is $ℤ$-linear in both arguments for every three objects $X$, $Y$, $Z$ of $\mathscr{C}$. This holds by definition of a preadditive category.

3. If $A$, $B$ and $C$ are three abelian groups and $β \colon A × B \to C$ is a $ℤ$-bilinear map, then $β(0, b) = 0$ for every element $b$ of $B$. For this equality we start with $$ β(0, b) = β(0 + 0, b) = β(0, b) + β(0, b) $$ and then subtract $β(0, b)$ from both sides of this overall equation. For the second equality we can proceed in the same way.

4. If $f \colon X \to Y$ is a morphism in $\mathscr{C}$, $Z$ is another object of $\mathscr{C}$ and $0_{Y, Z}$ is the zero object of the abelian group $\mathscr{C}(Y, Z)$, then the composite $0_{Y, Z} ∘ f$ is the zero object of the abelian group $\mathscr{C}(X, Z)$, i.e., $0_{Y, Z} ∘ f = 0_{X, Z}$. This is a combination of points 2 and 3. (We set $A ≔ \mathscr{C}(Y, Z)$, $B ≔ \mathscr{C}(X, Y)$, $C ≔ \mathscr{C}(X, Z)$ and $β(g, f) ≔ g ∘ f$.)

5. Every morphism $f \colon X \to \mathbf{0}$ in in $\mathscr{C}$ is already the zero element of the abelian group $\mathscr{C}(X, \mathbf{0})$. This holds because $$ f = 1_{\mathbf{0}} ∘ f = 0_{\mathbf{0}, \mathbf{0}} ∘ f = 0_{X, \mathbf{0}} $$ by definition of a category, point 1, and point 4 respectively.

6. As a consequence of point 5, there can exist at most one morphism from $X$ to $\mathbf{0}$ for every object $X$ of $\mathscr{C}$.

7. There exists for every object $X$ of $\mathscr{C}$ a morphism from $X$ to $\mathbf{0}$. This holds because we have the structure of an abelian group on $\mathscr{C}(X, \mathbf{0})$, and groups cannot be empty.

8. There exists for every object $X$ of $\mathscr{C}$ precisely one morphism from $X$ to $\mathbf{0}$. This is a combination of points 6 and 7.