# Affine maps and affine subspaces

In the Appendix of John M Lee's Textbook "Introduction to Smooth Manifolds" there is the following Exercise B.17

Suppose $$V$$ is a finite-dimensional vector space. Show that every affine subspace of $$V$$ is of the form $$F^{-1}(z)$$ for some affine map $$F:V \to W$$ and some $$z\in W$$.

Some definitions:

1. An affine subspace of $$V$$ is any subset of the form $$v+S=\{v+w:w\in S\}$$ with $$v\in V$$ and $$S$$ linear subspace of $$V$$.
2. If $$V$$ and $$W$$ are vector spaces, a map $$F:V\to W$$ is called an affine map if it can be written in the form $$F(v)=w+Tv$$ for some linear map $$T:V\to W$$ and some fixed $$w\in W$$.

Here is my argument

Let $$v_0+S$$ denote the generic affine subspace of $$V$$. Let $$\pi:V\to V/S$$ denote the canonical projection. Then $$\pi$$ is an affine map and $$\pi^{-1}(v_0+S)=v_0+S$$. So the thesis is true with $$W=V/S, F=\pi, z=v_0+S$$.

It seems correct to me but I don't use the hypothesis that $$V$$ is finite dimensional. So have I misunderstood the statement of the exercise or the proof is not correct?

• Perhaps you misunderstood how the author expected you to prove it. An introduction to smooth manifolds is not going to deal with infinite dimensional vector spaces much (probably not at all). So it is common to just toss out finite dimensionality a lot so as to avoid the issues that may arise with infinite dimensions. Rather than filter for situations where the assumption isn't needed, it is common just to make it always. Probably that is the case here. The author had a finite dimensional proof in mind, but the theorem isn't actually restricted to finite dimensions. – Paul Sinclair Apr 4 at 23:34