Dimension of preimage question

[ungaria]

let g be the inverse function of f (not necessarily a bijective inverse function)
If S is any subset of W then the pre-image of S is

g(S) = {v ε V: f(v) ε S}

Suppose that U is a subspace of W.
Prove that g(U) is a subspace of V.
Also prove that
dim( g(U) )= dim( U intersect Im(f) ) + dim( ker (f) )

For the first part,
by linearity, f(av1+v2)=af(v1)+f(v2)=au1+ u2 ε U where v1ε g(U), v2ε g(U)
so av1+v2 ε g(U)---> a g(u1)+g(u2)ε g(U)
if f(v)=v.then g(U) is not empty
therefore, g(U) is a subspace.

For the second part,
I am not sure. can i just simply assume that U=W? then g(U) = V and U intersect im(f) is im(f) so clearly dim(V)=dim imf + dim kerf

Could anyone verify my first proof and give me a hint for the second part? Thanks in advance.

 

[fresh_42]

Consider the restriction of $f$ on $g(U)$. Then $\operatorname{ker}f|_{g(U)}= \operatorname{ker}f$ and the rank-nullity-theorem gives $$
\dim(U \cap \operatorname{im}f) = \dim \operatorname{im}f|_{g(U)} = \dim g(U) - \dim \operatorname{ker}f|_{g(U)}=\dim g(U) - \dim \operatorname{ker}f
$$