Natural transformation

Remarks

The condition says that the naturality square with sides $F(f)$, $\alpha_B$, $\alpha_A$ and $G(f)$ is a commutative diagram. Example: on vector spaces over a field $K$, let $\mathrm{ev}_V \colon V \to V^{**}$ send $v$ to the map $\varphi \mapsto \varphi(v)$, where $V^*$ is the dual space. For a linear map $f \colon V \to W$ and $\psi \in W^*$, both $f^{**}(\mathrm{ev}_V(v))$ and $\mathrm{ev}_W(f(v))$ send $\psi$ to $\psi(f(v))$. So $\mathrm{ev}$ is a natural transformation from the identity functor to the double dual functor. On finite-dimensional spaces it is a natural isomorphism, by the canonical isomorphism with the double dual. Non-example: on the category of nonempty sets, choose for each set $X$ a point $x_X \in X$ and let $\alpha_X \colon X \to X$ be the constant map with value $x_X$. This family is not a natural transformation from the identity functor to itself. For a two-element set $Y$ and the constant map $f \colon X \to Y$ whose value is not $x_Y$, the condition would require $f(x_X) = x_Y$.