Why every object of a Galois category is the sum of its connected subobjects? Let $C$ be a Galois category.
    Let $X$ be an object of $C$.
    In Lenstra's notes on Galois theory for schemes, it is written that if $X$ is not connected then there is a subobject $Y \longrightarrow X$ with $\emptyset = F(0) \not= F(Y) \not= F(X)$.
    Using G3 one then finds $Z$ such that $X$ may be identified with $Y \coprod Z$ so that $F(X)$ is, by G5, equal to the disjoint union of $F(Y)$ and $F(Z)$.
    Arguing by induction on $\# F(X)$ one concludes that every object of $C$ is the sum of its connected subobjects (page 40).
I see why every object is the sum of connected subobjects but I don't understand why every object is the sum of all its connected subobjects.
 A: If I understand the question, the central claim is that any decomposition uses all the connected subobjects of $X$. In other words:

Claim: In a Galois category $(\mathcal{C}, F)$, given connected objects $B_1, B_2, \dots, B_k, B$ and a monomorphism $u: B \hookrightarrow B_1 \coprod B_2 \coprod \dots \coprod B_k=:X,$ $u$ factors for a unique $j$ as an isomorphism $u_j: B \stackrel{\simeq}\rightarrow B_j$ followed by the canonical map $B_j \rightarrow X$ (hence, $B$ and $B_j$ are equal as subobjects of $X$).

To show that, first berve the following

Observation: In a Galois category,  fibered products distribute over finite coproducts.

Proof of Observation: Enough to argue for binary coproducts. So consider $X \rightarrow Y, Z_1 \coprod Z_2\rightarrow Y$ in $\mathcal{C}$. We want to show that the canonical map $$can:(X\times_Y Z_1) \coprod (X\times_Y Z_2)\rightarrow X\times_Y (Z_1\coprod Z_2)$$ is an isomorphism. The fiber functor $F$ commutes with fiber products and finite coproducts (by (G4), (G5)), so in $\mathbf{Set}$ we obtain the same canonical map
$$F(can):(FX\times_{FY} FZ_1) \coprod (FX\times_{FY} FZ_2)\rightarrow FX\times_{FY} (FZ_1\coprod FZ_2)$$
which is an isomorphism of sets (this is a (trivial) property of $\mathbf{Set}$). By (G6), it follows that $can$ is itself an isomorphism. $\square$
Now the claim follows rather formally:
Proof of Claim: Given $u$ as in the claim, one can "refine B by the decomposition of $X$," i.e. look at
$$B=B \times_X X=B \times_X(B_1 \coprod B_2 \coprod \dots \coprod B_k)=(B\times_X B_1) \coprod (B \times_X B_2) \coprod \dots \coprod (B \times_X B_k).$$
Since $B$ is connected, it follows that one of the pieces $B \times_X B_j$ equals to $B$ as a subobject of $B$, an all the other are equal to $\emptyset$. That is, the canonical map $B\times_X B_j \rightarrow B$ is invertible, and $u_j$ is then obtained by considering its inverse followed by the second canonical map,
$$u_j: B \stackrel{\simeq}\leftarrow B\times_X B_j \hookrightarrow B_j.$$
(To see that $u_j$ is in fact an isomorphism, note that it is monic and $B, B_j$ are connected.) $\square$
