How to prove a set with at least one element is a separator?

Question

I'm reading Lawvere's Sets for Mathematics and got stuck at Exercise 1.15

In the category of abstract sets S, any set A with at least one element $1 \xrightarrow{x} A$ is a separator.

I can see that with the axiom that "the terminal object $1$ is a separator" and the following diagram, I ought to deduce the statement, but I cannot seem to write it down formally:

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Any hint please?

Answer 1

Let $X$ be any non-empty set. If $f,g:A\to B$ are two functions with $f\neq g$, then $f(a)\neq g(a)$ for some $a\in A$. Now choose an element $x\in X$ and define a function $h:X\to A$ such that $h(x)=a$. Then $fh\neq gh$.

Answer 2

Hint: what happens if $f_1 h=f_2 h$ for every $h:A\to X$? Particularly, if $f_1\neq f_2$ and $i:1\to X$ separates them, what about $i\circ !:A\to 1\to X$?

Answer 3

Definition. An object $S$ in a category $\mathcal{C}$ is a separator if and only if whenever

$$f_1,f_2:X\rightrightarrows Y$$ are arrows of $\mathcal{C}$ then $$(\forall x[S\overset{x} \rightarrow X \Rightarrow f_1 x = f_2 x])\Longrightarrow f_1 = f_2 $$ in other words $$f_1 \neq f_2 \Longrightarrow (\exists x[S\overset{x} \rightarrow X \hspace{0.3cm}\text{and}\hspace{0.3cm} f_1 x \neq f_2 x)$$

• THE TERMINAL OBJECT $1$ SEPARATES MAPPINGS IN $\mathcal{Set}.$

A one-element set $1$ is a separator in $\mathcal{S}$, i.e., if

$$f_1,f_2:X\rightrightarrows Y$$ then $$(\forall x[1\overset{x} \rightarrow X \Rightarrow f_1 x = f_2 x])\Longrightarrow f_1 = f_2 $$

in other words $$f_1 \neq f_2 \Longrightarrow (\exists x[1\overset{x} \rightarrow X \hspace{0.3cm}\text{and}\hspace{0.3cm} f_1 x \neq f_2 x)$$ Statement. In the category of abstract sets $\mathcal{Set}$, any set $A$ with at least one element $1\overset{x}\rightarrow A $ is also a separator.

Proof.

Suppose $f_1 \neq f_2$. Then $1$ separates them, there exists $1\overset{x} \rightarrow X$ such that $f_1 x \neq f_2 x$ $$1\overset{x}\rightarrow X\rightrightarrows Y$$ $A$ is a separator if and only if there exists $h:A\to X$ s.t. $f_1 h \neq f_2 h$. Consider $h = x\circ !$ $$A\overset{!}\to 1\overset{x}\rightarrow X\rightrightarrows Y$$ Let's show that $f_1 \circ x\circ !\neq f_2 \circ x\circ !$. Consider any element $1\overset{a}\to A$, it separates $f_1 \circ x\circ !$ and $ f_2 \circ x\circ !$ $$1\overset{a}\to A\overset{!}\to 1\overset{x}\rightarrow X\rightrightarrows Y$$ because $1\overset{a}\to A\overset{!}\to 1$ is $id_1$ identity arrow of $1$ $$f_1\circ x \circ ! \circ a = f_1\circ x \neq f_2\circ x = f_2\circ x \circ ! \circ a$$ Hence $$f_1 \circ x\circ !\neq f_2 \circ x\circ !$$ and $A$ is a separator.