The Stacks project

Proof. First assume that $x$ is either (1) initial or (2) final. In both cases, it follows that $\mathop{\mathrm{Mor}}\nolimits (x,x)$ is a trivial abelian group containing $\text{id}_ x$, thus $\text{id}_ x = 0$ in $\mathop{\mathrm{Mor}}\nolimits (x, x)$, which shows that each of (1) and (2) implies (3).

Now assume that $\text{id}_ x = 0$ in $\mathop{\mathrm{Mor}}\nolimits (x,x)$. Let $y$ be an arbitrary object of $\mathcal{A}$ and let $f \in \mathop{\mathrm{Mor}}\nolimits (x ,y)$. Denote $C : \mathop{\mathrm{Mor}}\nolimits (x,x) \times \mathop{\mathrm{Mor}}\nolimits (x,y) \to \mathop{\mathrm{Mor}}\nolimits (x,y)$ the composition map. Then $f = C(0, f)$ and since $C$ is bilinear we have $C(0, f) = 0$. Thus $f = 0$. Hence $x$ is initial in $\mathcal{A}$. A similar argument for $f \in \mathop{\mathrm{Mor}}\nolimits (y, x)$ can be used to show that $x$ is also final. Thus (3) implies both (1) and (2). $\square$

Proof. Suppose that $z = x \times y$ with projections $p : z \to x$ and $q : z \to y$. Denote $i : x \to z$ the morphism corresponding to $(1, 0)$. Denote $j : y \to z$ the morphism corresponding to $(0, 1)$. Thus we have the commutative diagram

where the diagonal compositions are zero. It follows that $i \circ p + j \circ q : z \to z$ is the identity since it is a morphism which upon composing with $p$ gives $p$ and upon composing with $q$ gives $q$. Suppose given morphisms $a : x \to w$ and $b : y \to w$. Then we can form the map $a \circ p + b \circ q : z \to w$. In this way we get a bijection $\mathop{\mathrm{Mor}}\nolimits (z, w) = \mathop{\mathrm{Mor}}\nolimits (x, w) \times \mathop{\mathrm{Mor}}\nolimits (y, w)$ which show that $z = x \amalg y$.

We leave it to the reader to construct the morphisms $p, q$ given a coproduct $x \amalg y$ instead of a product. $\square$

Proof. Suppose $F$ is additive. A direct sum $z$ of $x$ and $y$ is characterized by having morphisms $i : x \to z$, $j : y \to z$, $p : z \to x$ and $q : z \to y$ such that $p \circ i = \text{id}_ x$, $q \circ j = \text{id}_ y$, $p \circ j = 0$, $q \circ i = 0$ and $i \circ p + j \circ q = \text{id}_ z$, according to Remark 12.3.6. Clearly $F(x), F(y), F(z)$ and the morphisms $F(i), F(j), F(p), F(q)$ satisfy exactly the same relations (by additivity) and we see that $F(z)$ is a direct sum of $F(x)$ and $F(y)$. Hence, $F$ transforms direct sums to direct sums.

To see that $F$ transforms zero to zero, use the characterization (3) of the zero object in Lemma 12.3.2. $\square$

Proof. Let $f : z' \to x \oplus y$ be a morphism such that $q \circ f = 0$. We have to show that there exists a unique morphism $g : z' \to x$ such that $f = i \circ g$. Since $i \circ p + j \circ q$ is the identity on $x \oplus y$ we see that

and hence $g = p \circ f$ works. Uniqueness holds because $p \circ i$ is the identity on $x$. The proof of the second statement is dual. $\square$

Proof. Part (1) follows easily from the uniqueness required in the definition of a kernel. The proof of (2) is dual. Part (3) follows from (2), since the coimage is a cokernel. Similarly, (4) follows from (1). $\square$

Proof. There is a canonical morphism $\mathop{\mathrm{Coim}}(f) \to y$ because $\mathop{\mathrm{Ker}}(f) \to x \to y$ is zero. The composition $\mathop{\mathrm{Coim}}(f) \to y \to \mathop{\mathrm{Coker}}(f)$ is zero, because it is the unique morphism which gives rise to the morphism $x \to y \to \mathop{\mathrm{Coker}}(f)$ which is zero (the uniqueness follows from Lemma 12.3.11 (3)). Hence $\mathop{\mathrm{Coim}}(f) \to y$ factors uniquely through $\mathop{\mathrm{Im}}(f) \to y$, which gives us the desired map. $\square$