Composition of nonlinear maps

Question

Let: $$f: V \to W$$ and $$g: W \to U$$ where V, W, and U are some vector spaces. Can you please give an example (an interesting one, if possible) where $$g \circ f$$ is a homomorphism where f or g are not homomorphisms, if such a thing exists. (*Edit: I apologize for my impreciseness, but here I implicitly assumed, hence the "interesting", that g o f was not necessarily an isomorphism between V and U i.e. that f and g were not necessarily bijective so that the resulting would not be such a triviality as the identity function - it can be an isomorphism so long as it is nontrivial)

In other words, I proved that a composition of linear maps is linear, but am asking for a nontrivial counterexample of the converse.

In particular, I was somewhat interested in whether a function g can be constructed such that: $$f:x \mapsto x^2$$ $$g(f(x+y))=g(f(x))+g(f(y))$$ where we take domain and codomain of f and g to be the reals.

Answer 1

A function $g$ like you wish can be given, but it is not "interesting", as far as I'm concerned. Note that a nonzero linear function $\mathbf R \to \mathbf R$ is one-to-one, as $f$ is not one-to-one, $g$ has to be zero on the image of $f$. But on $(-\infty,0)$ you can take $g$ to be what you like. That is for example, define $g \colon \mathbf R \to \mathbf R$ by $$ g \colon x \mapsto \begin{cases} 0 & x \ge 0\\ \sin x & x < 0 \end{cases} $$ then $g \circ f = 0$ is linear.

Answer 2

It's not very exciting but in general if $f:V \rightarrow W$ is nonlinear but invertible then $f\circ f^{-1} = Id_V$ is linear.

Answer 3

The only way you can make your example work is to take $g(y)=0$ for all $y\geq0$ so that $g\circ f$ is the zero map $\Bbb R\to\Bbb R$. You can however define $g(y)$ to be whatever you like for $y<0$; in particular you can very well make $g$ not linear (nor injective, surjective, or most any nice property you would like to avoid).

The reason that you choices are so limited, is that $g\circ f$ has to be a linear map, which because of $f(x)=f(-x)$ takes the same value for any $x\in\Bbb R$ as it takes for $-x$, and the zero map is the only such map (proof of which I leave as exercise).