A different way to define homomorphism.

Question

Let $V(K)$ and $W(k)$ be two vector space, and $f:V \to W$ an homomorphism.

Then $f$ maps subspaces in subspaces, forwards and backwards: $$A \in \Sigma[V] \Longrightarrow f(A)\in \Sigma[W]$$ $$A \in \Sigma[W] \Longrightarrow f^{-1}(A)\in \Sigma[V]$$ My question is: do these conditions fully characterize homomorphisms ? Otherwise stated, do they imply that $f \in Hom(V,W)$ ?

That's my try to prove the thesis.

Let $$[v_1, ... , v_n]$$ a set of vectors in $V$ . To prove that $$f(k^{p}v_{p})=h^{p}f(v_{p})$$ consider $$M:=<[v_1, ... , v_n]>$$ $$N:=<[f(v_1), ... , f(v_n)]>$$ $$T:=f(M)$$ $$S:=f^{-1}(N)$$

Obviously $$f(k^{p}v_{p}) \in T$$ From the first hypothesis $$h^{p}f(v_{p}) \in T$$ Now, again from the fist hypotesis, $$\Delta := f(k^{p}v_{p})- h^{p}f(v_{p}) \in T$$ and I would like, using the second one, prove that $\Delta = 0$ . I'm grateful for your attention and I will greatly esteem any hint!

Answer 1

If $V$ and $W$ are both $1$-dimensional, e.g. $V=W=k$ then the only subspaces are $\{0\}\subset k$ and $k \subseteq k$. Thus your conditions on a function $f\colon V\to W$ translate to:

1. $f(0)=0$ (since $f(\{0\})=k$ is impossible)
2. $f^{-1}(\{0\})=\{0\}$ or $f^{-1}(\{0\})=k$
3. $f(k)=k$ or $f(k)=\{0\}$
4. $f^{-1}(k)=k$ (since $f^{-1}(k)=\{0\}$ is impossible)

These are precisely the function $f\colon k\to k$ such that

• $f=0$ or
• $f(0)=0$ and $f\colon k\setminus \{0\}\to k\setminus\{0\}$ is any surjective function

Clearly not every such function is a homomorphism (unless $k=\mathbb{F}_2$).