$V/(U \cap W)$ is finite dimensional iff $V/W$ and $V/U$ are finite dimensional

Question

Where obviously $U, W$ are subspaces of a vector space $V$. I have one direction, i.e. if $V/W$ and $V/U$ are finite dimensional, $V/(U\cap W)$ is finite dimensional using the isomorphism theorems. How can I get the other direction?

Answer 1

HINT: Show that $V/U$ is a quotient of $V/(U\cap W)$.

Answer 2

$\Leftarrow$ Assume dim$(V/W)<\infty$ and dim$(V/U)<\infty$. We know then dim$(V/W \oplus V/U) = (\text{dim}(V/W)\cdot \text{dim}(V/U))<\infty$. We know there is an injective function from $V/(W\cap U) \xrightarrow{\phi} V/U\oplus V/W$, and if dim$(V/(W\cap U))\not < \infty$ then there could not exist an injective linear transformation between $V/(W\cap U)$ and $V/U \oplus V/W$. So then dim$(V/(W\cap U))<\infty$.

(The existence of that injective transformation was given in an earlier problem about short exact sequences.)