In a concrete category, every section is an injective morphism, but the converse does not hold. For example, in the category of groups, the homomorphism $f:\mathbb Z\to\mathbb Z$ given by $f(n)=2n$ is injective, but there is not a homomorphism $g:\mathbb Z\to\mathbb Z$ such that $g\circ f=\operatorname{id}_\mathbb Z$.
Similarly, every retraction is a surjective morphism. I believe the converse does not hold – that is, there is a surjective morphism which is not a retraction. Is this true?
On
The category of topological spaces and continuous maps provides nice examples.
Let $X$ be a set with more than one element. Giving $X$ the discrete topopogy yields a space $X^d$ and giving it the trivial topology a space $X^t$. The identity $I : X^d \to X^t$ is a continuous bijection, but it is neither a section nor a retracton. In fact, the only function $f : X^t \to X^d$ such that $f \circ I = id_{X^d}$ resp. $I \circ f = id_{Z^t}$ on the level of sets is the identity which is not continuous.
On
Let $G$ be any nontrivial group, and consider the concrete category of $G$-sets. Then the unique function $G \to \{ * \}$ is a surjective homomorphism, where the action of $G$ on $G$ is by left multiplication, and the action of $G$ on $\{ * \}$ is trivial. However, this does not have any section $\{ * \} \to G$; in fact, there are no $G$-equivariant functions $\{ * \} \to G$ at all.
In the category of groups, the homomorphism $q : \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ sending each integer to its parity is a surjective homomorphism, but there is no homomorphism $s : \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}$ such that $q \cdot s = \mathrm{id}_{\mathbb{Z}/2\mathbb{Z}}$. Suppose there were. Since $1 + 1 = 0$ in $\mathbb{Z}/2\mathbb{Z}$, we would have $s(1) + s(1) = 0$, implying that $s(1) = 0 = s(0)$. So $s$ isn't injective, contradicting the fact that it's a section.