Why Is There an Isomorphism on Sets If They Lack Algebraic Structure?

[kntsy]

Hi i just start learning algebra.
Here are some definitions and examples given in Wikipedia:
1.An isomorphism is a bijective map f such that both f and its inverse $f^{-1}$ are homomorphisms, i.e., structure-preserving mappings.
2.A homomorphism is a structure-preserving map between two algebraic structures (such as groups, rings, or vector spaces).
3.Consider the logarithm function: For any fixed base b, the logarithm function $\log_b$ maps from the positive real numbers $\mathbb{R^+}$ onto the real numbers $\mathbb{R}$; formally:$\log_{b}:\mathbb{R^+}\rightarrow\mathbb{R}$
This mapping is one-to-one and onto, that is, it is a bijection from the domain to the codomain of the logarithm function. This is an isomorphism on set.

Question(not h.w.): Set is not an algebraic structure as no operation is defined in it. From(1), isomorphism is also homoporhism.
From(2), there is no homoporhism on set.
So why is there isomorphism on set?

 

[Hurkyl]

Actually, a set is an algebraic structure. It's just that this particular structure doesn't have any constant symbols or any nontrivial operations, and doesn't have any nontrivial equational axioms.

So, every function is a homomorphism.



By "trivial operation" I mean things like projections L(x,y) --> x (L for "left"). And by "trivial axiom" I mean the relations between trivial operations -- e.g. L(L(x,y),z) = L(x,L(y,z)), or x=x.