let g be the inverse function of f (not necessarily a bijective inverse function)(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Dimension of preimage question!

Can you offer guidance or do you also need help?

**Physics Forums | Science Articles, Homework Help, Discussion**