vector space monomorphism splits

Proposition

Let \(K\) be a field, \(U,V\) vector spaces and \(f: U \hookrightarrow V\) be a vectorspace-monomorphism
Then \(f\) is split mono, i.e. there exists a vector space homomorphism \(\pi: V \rightarrow U\) such that

\begin{align*} \pi \circ f = \mathrm{id}_U \end{align*}

Proof

Choose a basis \(u_i\) of \(U\) (cf. Existenz einer Basis in einem Vektorraum).
Then since \(f: U \twoheadrightarrow V\) is injective, we may conclude that \(f(u_i)\) is linearly independent in \(V\) (cf. Vektorraummonomorphismus erhält lineare Unabhängigkeit).
Thus we may extend \(f(u_i)\) to a basis (cf. erweiterung einer linear unabhängigen Menge zu einer Basis) \(f(u_i),v_j\)

Then define \(\pi\) via.

\begin{align*} \pi: V \rightarrow& U \\ f(u_i) \mapsto& u_i v_j \mapsto& 0 \end{align*}

and Existenz und Eindeutigkeit eines Homomorphismus durch die Abbildungen der Basis.

Note that \(f(u_i) = f(u_j) \Rightarrow u_i = u_j\) (as \(f\) is injective) which shows that the assignment

\begin{align*} f(u_i) \mapsto& u_i \end{align*}

is welldefiend

It follows that

\begin{align*} \pi(f(u_i)) = u_i \end{align*}

hence \(\pi \circ f\) is the identity on the basis \(u_i\).

Hence by uniqueness of the extension, it is the identity on \(U\)

Date: nil

Author: Anton Zakrewski

Created: 2025-01-15 Mi 07:47