I showed $V$ is isomorphic to $U \times (V/U)$ without using the assumption that $V/U$ is finite-dimensional. What did I do wrong?

Question

Exercise.

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

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Outline of my proof.

We need to construct a linear bijection between $V$ and $U \times (V/U)$, where $U \times (V/U) = \{(u, v+U) : u \in U, v \in V\}$.

I defined a mapping $T: U \times (V/U) \to V$ by

$$ T(u,v+U) = v+u $$

where $v+u \in V$. I was then able to show the following are true about $T$:

1. Linear
2. Injective
3. Surjective

Question. In my full proof, I didn't use the fact that $V/U$ is finite-dimensional, which makes me think something is wrong. Is it because my $T$ is not well-defined? How come I was able to show the above three properties?

Answer 1

The issue is your "function" $T$ is not well-defined. In particular, it is possible that $a+U = b+U$ and $a \neq b$. In this case, you would have $T(0,a+U)=a$ and $T(0,a+U)=T(0,b+U)=b$, which is contradictory to being a function.