Is the following argument correct?

Suppose $U$ is a subspace of $V$ such that $V/U$ is finite dimensional. Prove that $V$ is isomorphic to $U\times (V/U)$.

Proof. Let $v_1+U,v_2+U,\dots,v_n+U$ be a basis of $V/U$ and define the map $\Delta:U\times (V/U)\to V$ such that $\displaystyle\Delta\left(u,\sum_{k=1}^{n}\lambda_k(v_k+U)\right) = u+\sum_{k=1}^{n}\lambda_kv_k$.

We leave the proof that $\Delta$ is a linear map to the reader and instead proceed by proving that $\Delta:U\times (V/U)\to V$ is both injective and surjective.

Assume $\displaystyle\Delta\left(u_1,\sum_{k=1}^{n}\lambda_k(v_k+U)\right) = \Delta\left(u_2,\sum_{k=1}^{n}\alpha_k(v_k+U)\right)$ therefore $\displaystyle u_1+\sum_{k=1}^{n}\lambda_kv_k = u_2+\sum_{k=1}^{n}\alpha_kv_k$ equivalently $\displaystyle u_1-u_2 = \sum_{k=1}^{n}(\alpha_k-\lambda_k)v_k$ but then $\displaystyle\left(\sum_{k=1}^{n}(\alpha_k-\lambda_k\right)v_k)\in U$ which by theorem $\textbf{3.85}$ is equivalent to $\displaystyle\sum_{k=1}^{n}\alpha_kv_k+U = \sum_{k=1}^{n}\lambda_kv_k+U$, and so $\displaystyle\sum_{k=1}^{n}\alpha_k(v_k+U) = \sum_{k=1}^{n}\lambda_k(v_k+U)$ but since $v_1+U,v_2+U,\dots,v_n+U$ is a basis of $V/U$ it follows that $\lambda_k = \alpha_k,\forall k$ and so $u_1 = u_2$. Thus the map in question is injective.

Now let $v\in V$, thus the affine subset $v+U\in V/U$ and thus for some scalars $\lambda_1,\lambda_2,\dots,\lambda_n\in\mathbf{F}$ we have $\displaystyle v+U = \sum_{k=1}^{n}\lambda_k(v_k+U)$, equivalently by theorem $\textbf{3.85}$ we have $\displaystyle \left(v - \sum_{k=1}^{n}\lambda_kv_k\right)\in U$, then given the above definition of $\Delta$ is it evident that $\displaystyle\Delta\left(v-\sum_{k=1}^{n}\lambda_kv_k,\sum_{k=1}^{n}\lambda_k(v_k+U)\right) = v$.


NOTE: Theorem $\textbf{3.85}$ is as follows Given a subspace $U$ of the vector space $V$ and vectors $v,w\in V$ the following are equivalent.

  • $v-w\in U$

  • $v+U = w+U$

  • $(v+U)\cap(w+U)\neq\varnothing$

  • $\begingroup$ what does theorem 3.85 state? $\endgroup$ – Alvin Lepik Sep 5 '18 at 14:05
  • $\begingroup$ @AlvinLepik I have added theorem 3.85 now sorry for not mentioning it earlier. $\endgroup$ – Atif Farooq Sep 5 '18 at 14:10

Your proof is correct.

Note that you have implicitly chosen (arbitrary) preimages $v_i$ of the basis elements of $V/U$.
Now let $W:=\mathrm{span}(v_1,\dots v_n)\, \le U$.

Then we have $W\cong U/V$ thus $U\times (V/U) \cong U\times W$, and also $U\times W\cong U\oplus W=V$.

  • $\begingroup$ WoW so many isomorphisms, funny enough the subsequent exercises in my book go on to establish these very results. Thanks for the verfication. $\endgroup$ – Atif Farooq Sep 5 '18 at 18:08

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