Lemma 58.3.7. Let $(\mathcal{C}, F)$ be a Galois category. Let $X \to Y \in \text{Arrows}(\mathcal{C})$. Then

1. $F$ is faithful,

2. $X \to Y$ is a monomorphism $\Leftrightarrow F(X) \to F(Y)$ is injective,

3. $X \to Y$ is an epimorphism $\Leftrightarrow F(X) \to F(Y)$ is surjective,

4. an object $A$ of $\mathcal{C}$ is initial if and only if $F(A) = \emptyset$,

5. an object $Z$ of $\mathcal{C}$ is final if and only if $F(Z)$ is a singleton,

6. if $X$ and $Y$ are connected, then $X \to Y$ is an epimorphism,

7. if $X$ is connected and $a, b : X \to Y$ are two morphisms then $a = b$ as soon as $F(a)$ and $F(b)$ agree on one element of $F(X)$,

8. if $X = \coprod _{i = 1, \ldots , n} X_ i$ and $Y = \coprod _{j = 1, \ldots , m} Y_ j$ where $X_ i$, $Y_ j$ are connected, then there is map $\alpha : \{ 1, \ldots , n\} \to \{ 1, \ldots , m\}$ such that $X \to Y$ comes from a collection of morphisms $X_ i \to Y_{\alpha (i)}$.

Proof. Proof of (1). Suppose $a, b : X \to Y$ with $F(a) = F(b)$. Let $E$ be the equalizer of $a$ and $b$. Then $F(E) = F(X)$ and we see that $E = X$ because $F$ reflects isomorphisms.

Proof of (2). This is true because $F$ turns the morphism $X \to X \times _ Y X$ into the map $F(X) \to F(X) \times _{F(Y)} F(X)$ and $F$ reflects isomorphisms.

Proof of (3). This is true because $F$ turns the morphism $Y \amalg _ X Y \to Y$ into the map $F(Y) \amalg _{F(X)} F(Y) \to F(Y)$ and $F$ reflects isomorphisms.

Proof of (4). There exists an initial object $A$ and certainly $F(A) = \emptyset$. On the other hand, if $X$ is an object with $F(X) = \emptyset$, then the unique map $A \to X$ induces a bijection $F(A) \to F(X)$ and hence $A \to X$ is an isomorphism.

Proof of (5). There exists a final object $Z$ and certainly $F(Z)$ is a singleton. On the other hand, if $X$ is an object with $F(X)$ a singleton, then the unique map $X \to Z$ induces a bijection $F(X) \to F(Z)$ and hence $X \to Z$ is an isomorphism.

Proof of (6). The equalizer $E$ of the two maps $Y \to Y \amalg _ X Y$ is not an initial object of $\mathcal{C}$ because $X \to Y$ factors through $E$ and $F(X) \not= \emptyset$. Hence $E = Y$ and we conclude.

Proof of (7). The equalizer $E$ of $a$ and $b$ comes with a monomorphism $E \to X$ and $F(E) \subset F(X)$ is the set of elements where $F(a)$ and $F(b)$ agree. To finish use that either $E$ is initial or $E = X$.

Proof of (8). For each $i, j$ we see that $E_{ij} = X_ i \times _ Y Y_ j$ is either initial or equal to $X_ i$. Picking $s \in F(X_ i)$ we see that $E_{ij} = X_ i$ if and only if $s$ maps to an element of $F(Y_ j) \subset F(Y)$, hence this happens for a unique $j = \alpha (i)$. $\square$

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