# Why is an isomorphism between (total) algebras required to have an inverse which is a homomorphism?

Let us consider homomorphisms between partial algebras as defined in

https://planetmath.org/homomorphismbetweenpartialalgebras

There, an isomorphism from $$A$$ to $$B$$ is a

1. bijective homomorphism from $$A$$ to $$B$$

2. such that its inverse is a homomorphism, too.

I can see that for partial algebras the second condition is necessary. Is the second condition also necessary if $$A$$ is a total algebra, or can this condition be dropped? I.e., is condition 2 implied by condition 1 when $$A$$ is total?

• The second condition is always necessary, but for some structures it is implied by the first. Is your question if 2 is implied by 1? – Dave Nov 9 '18 at 2:34

Yes, the second condition is implied by the first when $$A$$ is total. Indeed, if $$f$$ is a bijective (or even just surjective) homomorphism $$A\to B$$ and $$A$$ is total, then $$B$$ is also total, since for all function symbol $$\varphi$$ of arity $$n$$ and all $$b_1,\cdots,b_n\in B$$ we can find $$a_1,\cdots,a_n\in A$$ such that $$f(a_i)=b_i$$, and then $$\varphi(b_1,\cdots, b_n)=\varphi(f(a_1),\cdots,f(a_n))$$ is defined and equal to $$f(\varphi(a_1,\cdots,a_n))$$.
In particular $$f$$ is a bijective homomorphism between algebras, and thus its inverse is also an homomorphism by the usual argument.