Theorem: Let $U : \mathbf{A} \rightarrow \mathbf{Set}$ have a left-adjoint $F \dashv U$. Suppose that for at least one $A \in \mathbf{A}$, $U(A)$ has more than one element. Then for each $S$, the unit $\eta_S : S \rightarrow UF(S)$ is an injective map of sets.

Proof: We will first establish that, under the given hypothesis, $F$ must be faithful. To see this, suppose we have $g, h: S \rightarrow T$ such that $g \neq h$. The goal is to show that $F(g) \neq F(h)$. Indeed, if $g \neq h$, there exists some $x \in S$ such that $g(x) \neq h(x)$. Now, let $A \in \mathbf{A}$ be such that $\abs{U(A)} > 1$, and let $a, b \in U(A)$ be elements. Define $f : T \rightarrow U(A)$ with the property $f(g(x)) = a$ and $f(h(x)) = b$. Then, clearly, $f \circ g \neq f \circ h$. But, $f \circ g, f \circ h : S \rightarrow U(A)$ each have a transpose $F(S) \rightarrow A$. In particular, that transpose can be given explicitly as:

$$\overline{f \circ g} = \varepsilon_A \circ F(f \circ g) = \varepsilon_A \circ F(f) \circ F(g) = \overline{f} \circ F(g)$$

and similarly $\overline{f \circ h} = \overline{f} \circ F(h)$. Since $f \circ g \neq f \circ h$, their transposes are not equal either, and $\overline{f} \circ F(g) \neq \overline{f} \circ F(h)$, which means $F(g) \neq F(h)$. So $F$ is faithful.

We now show that faithfulness of $F$ implies the desired property of the unit. Fix an $S \in \mathbf{Set}$. Saying that $\eta_S$ is injective amounts to saying that, for any $g, h: S' \rightarrow S$, $\eta_S \circ g = \eta_S \circ h$ implies $g = h$. So, lets assume we have such a $g$ and $h$ with $\eta_S \circ g = \eta_S \circ h$. Note that $\eta_S \circ g$ is an arrow $S' \rightarrow U(F(S))$ in $\mathbf{Set}$, and as such it has a transpose $\overline{\eta_S \circ g} : F(S') \rightarrow F(S)$. Furthermore, this transpose can be calculated explicitly as:

$$\overline{\eta_S \circ g} = \varepsilon_{F(S)} \circ F(\eta_S \circ g) = \varepsilon_{F(S)} \circ F(\eta_S) \circ F(g) = F(g)$$

where the last equality follows precisely due to the triangle identity, which says that $\varepsilon_{F(S)} \circ F(\eta_S) = 1_{F(S)}$. Similarly we obtain $\overline{\eta_S \circ h} = F(h)$. Then, since $\eta_S \circ g = \eta_S \circ h$, their transposes are equal, and we have $F(g) = F(h)$. Since $F$ is faithful, we have $g = h$. So, $\eta_S$ is injective.