Let $(\mathcal{C},U)$ be concrete category over $\mathcal{SET}$ and let $\alpha:X\to Y$ be a morphism in $\mathcal{C}$ .

Definition. Morphism $\alpha$ is called injective (resp. surjective) if $U(\alpha)$ is an injection (resp. surjection).

Some properties of injective and surjective morphisms:

Proposition. Assume that $\alpha:X\to Y$ is a morphism in $\mathcal{C}$ .
$\mathrm{i)}$ If $\alpha$ is injective (resp. surjective), then $\alpha$ is a monomorphism (resp. epimorphism);
$\mathrm{ii)}$ If $\alpha$ is a section (resp. retraction), then $\alpha$ is injective (resp. surjective).

Proof. $\mathrm{i)}$ Assume that $\alpha$ is injective and let $\beta_1,\beta_2:Z\to X$ be morphisms in $\mathcal{C}$ such that $\alpha\circ\beta_1=\alpha\circ\beta_2$ . Then $$U(\alpha\circ\beta_1)=U(\alpha\circ\beta_2)$$ and this implies that $$U(\alpha)\circ U(\beta_1)=U(\alpha)\circ U(\beta_1).$$ Since $U(\alpha)$ is injective, we obtain that $U(\beta_1)=U(\beta_2)$ and since $U$ is faithful, we get that $$\beta_1=\beta_2.$$ Analogously we prove, that surjective morphism is an epimorphism.

$\mathrm{ii)}$ . Assume that $\alpha:X\to Y$ is a section. Then there exists $\beta:Y\to X$ such that $\beta\circ\alpha=\mathrm{id}_{X}$ . Thus we have $$U(\beta)\circ U(\alpha)=\mathrm{id}_{U(X)},$$ so $U(\alpha)$ is injective. Analogously retractions are surjective. $\square$