I'm looking through some notes on algebra isomorphisms and have encountered two slightly different notions of isomorphisms, depending on whether the algebras are Banach or not. All algebras are assumed to be complex.
Let $A,B$ be algebras over $\mathbb{C}$. We say that $A, B$ are isomorphic if there is a bijection $F:A\rightarrow B$ such that $\forall\ {x,y\in A};\ \forall\ \lambda,\mu\in\mathbb C:$ $$F(\lambda x+\mu y)=\lambda F(x)+\mu F(y);\ F(xy)=F(x)F(y).$$
In the case that $A,B$ are Banach algebras, then we require $F:A\rightarrow B$ to be a bounded linear operator which is invertible, and satisfies $$F(xy)=F(x)F(y);\ \forall_{x,y\in A}.$$
Since $F$ is invertible, (I think, at least) then we are ensured that $F$ is bijective; hence we 'hit' the definition for non-Banach algebras. I'm wondering why we require boundedness in the case of Banach algebras.