Let $V$ be a finite-dimensional vector space and $W,W_1,W_2$ subspaces of $V$, such that $V=W_1\oplus W_2$, $W \cap W_2=\{0\}$ and $\dim W= \dim W_1$. Prove that there exists a linear transformation $ f:W_1\to W_2$, such that $$W=\{v\in V \mid \exists w_1\in W_1:v=w_1+f(w_1)\}$$

My work so far:

Using that $\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2)$, for subspaces $U_1,U_2$ of $V$, we get that $V=W\oplus W_2$ and now my aim was to find an explicit linear transformation which has the required properties but I haven't been successful with this so far.

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    $\begingroup$ Do you require $f$ to be a linear transformation? $\endgroup$ – Aryaman Maithani Jan 2 at 17:46
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    $\begingroup$ @AryamanMaithani Oh yes I do, thank you, I have edited the question. $\endgroup$ – user Jan 2 at 17:56

Let us construct a function $f:W_1 \to W_2$ as follows.
Given any $w \in W$, it has a unique representation as $$w = w_1 + w_2,$$ where $w_1 \in W_1$ and $w_2 \in W_2$.
We define $f(w_1) = w_2$. It is clear that this $f$ does actually satisfy the condition you want. That is, $W = \{w_1 + f(w_1) \mid w_1 \in W_1\}$.
However, what is still left is to show that $f$ is indeed a well-defined linear function.

First, we show given any $w_1 \in W,$ there do exist vectors $w \in W$ and $w_2 \in W_2$ such that $w = w_1 + w_2$, that is, $f(w_1)$ has some value.
This is easy as $w_1 \in W_1 \subset V = W \oplus W_2.$ Thus, $w_1 = w + w_2$ for some $w \in W$ and some $w_2 \in W_2$. Rearranging the equation and using the fact that $-w_2 \in W_2$ gives us the desired result.

Secondly, we show that there is no ambiguity with this choice of $f(w_1).$
Suppose that $w_1 + w_2 = w \in W \ni \widehat{w} = w_1 + \widehat{w_2}$ for some $w_2, \widehat{w_2} \in W_2.$ We want to show that $w_2 = \widehat{w_2}.$
Note that $w - \widehat{w} = w_2 - \widehat{w_2}.$
The LHS is an element of $W$ and the RHS of $W_2.$ Thus, we get that $w_2 - \widehat{w_2} \in W \cap W_2 = \{0\}$, that is, $w_2 = \widehat{w_2}$, as desired.

Thirdly, we show that $f$ is indeed a linear map.
Suppose $x, y \in W_1$ and $\alpha \in \mathbb{F}$, the field over which $V$ is a vector space.
By construction, we have that $\alpha x + f(\alpha x) = X \in W$ and $y + f(y) = Y \in W$.
Thus, $\underbrace{\alpha x + y}_{\in W_1} + \underbrace{f(\alpha x) + f(y)}_{\in W_2} = \underbrace{X+Y}_{\in W}.$
By our definition, we get that $f(\alpha x + y) = f(\alpha x) + f(y)$ and hence, $f$ is a linear map.



From the hypothesis we have $V=W_1\oplus W_2=W\oplus W_2$ so for every $v\in V$ we can write by uniquely way: $$v=w_1+w_2=w+\omega_2$$ where $w_1\in W_1, w\in W$ and $w_2,\omega_2\in W_2$. Hence we have

$$w=w_1+ \underbrace{(w_2-\omega_2)}_{\in W_2}$$ Now we construct the function $f:W_1\to W_2, w_1\mapsto w_2-\omega_2$ and we prove that $f$ is a linear transformation and we are done.

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    $\begingroup$ Thank you very much for the answer but don't we have the problem that we can find $\hat{v}\neq v\in V$, such that $\hat{v}=w_1+\hat{w_2}$ and then $f$ would not be well-defined, would it? $\endgroup$ – user Jan 2 at 18:58

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