Suppose $V$ is a vector space and subspaces $U_1,U_2$ of $V$ are such that $U_1 \times U_2$ is isomorphic to $U_1+U_2$ but $U_1+U_2$ is not a direct sum. Prove that $V$ must be infinite dimensional.
Attempt: We are given that $\gamma:U_1\times U_2 \rightarrow U_1+U_2$ is an isomorphism.
Therefore, $\gamma$ is injective and surjective and thus for any $u_1+u_2 \in U_1+U_2,$ there exist a unique tuple $(u_1,u_2) \in U_1 \times U_2$
Then, $U_1+U_2$ should be a direct sum irrespective of whether $V$ is infinite-dimensional or not. ( since $\gamma$ is injective, for any $u_1+u_2 \in U_1+U_2$, there should exists a unique tuple $(u_1,u_2) \in U_1 \times U_2 \implies$ there's a unique way of expressing $u_1 +u_2 \implies U_! + U_2$ must be a direct sum )
Could someone please point-out where am I making a mistake? Thanks a lot!
This is vague. You seem to be mixing up "1 unique way of expressing $y \in U_1 + U_2$ as $\gamma(x) = y$" with "1 unique way of expressing $y \in U_1 + U_2$ as $y = u_1 + u_2$ where $u_i \in U_i$."
That $\gamma$ is an isomorphism gives the former interpretation. That $U_1+U_2$ is a direct sum requires the second interpretation. Therefore, regardless of how you formalize that vague statement, you'll end up using an implication that is false.
With your new edit, you are begging the question when you say
This notation assumes you have $\gamma(u_1,u_2) = u_1+u_2,$ which is a pretty strong assumption (it in fact would imply $U_1+U_2$ is a direct sum). Instead, all you have is that given any $u_1 + u_2 \in U_1 + U_2$, we have a unique $(v_1, v_2) \in U_1\times U_2$ such that $\gamma(v_1,v_2) = u_1+u_2.$
In answer to
Assuming $V$ is finite dimensional implies $U_1, U_2, U_1\times U_2, U_1+U_2$ are all finite dimensional. Now consider $\gamma : U_1 \times U_2 \to U_1 +U_2$ defined by $\gamma(u_1,u_2) = u_1 + u_2.$ This is obviously a surjective linear map, and by using the rank-nullity theorem, we can show $\text{Nullity } \gamma = 0$, so $\gamma$ is an isomorphism and $U_1+U_2$ is a direct sum.