On $\lambda$-homomorphic skew braces

For a skew left brace $(G,\cdot ,\circ )$, the map $\lambda : (G, \circ) \to $\mathrm{A}\mathrm{u}\mathrm{t}$ ;(G, \cdot),~~a \mapsto \lambda_a$,where$\lambda_a(b) = a^{-1} \cdot (a \circ b)$forall$a, b \in G$,isagrouphomomorphism\mathrm{.}Then$\lambda$canalsobeviewedasamapfrom$(G, \cdot)$to$$\mathrm{A}\mathrm{u}\mathrm{t}$; (G, \cdot)$, which, in general, may not be a homomorphism. We study skew left braces $(G,\cdot ,\circ )$ for which $\lambda : (G, \cdot) \to $\mathrm{A}\mathrm{u}\mathrm{t}$; (G, \cdot)$isahomomorphism\mathrm{.}Suchskewleftbraceswillbecalled$\lambda$-homomorphic. We formulate necessary and sufficient conditions under which a given homomorphism $\lambda :(G,\cdot )\to \mathrm{A}\mathrm{u}\mathrm{t}(G,\cdot )$ gives rise to a skew left brace, which, indeed, is $\lambda$-homomorphic. As an application, we construct skew left braces when $(G,\cdot )$ is either a free group or a free abelian group. We prove that any $\lambda$-homomorphic skew left brace is an extension of a trivial skew brace by a trivial skew brace. Special emphasis is given on $\lambda$-homomorphic skew left brace for which the image of $\lambda$ is cyclic. A complete characterization of such skew left braces on the free abelian group of rank two is obtained.