Direct sum of two vector spaces

Question

Can someone point out where is the fallacy in my proof below?

Claim: If $U_1,U_2,W$ are subspaces of a vector space $V$ and $V = U_1\bigoplus W = U_2 \bigoplus W $, then $U_1 = U_2$

My counter example: Let $V = R^2$, the 2-dimensional space.

$W$: the x-axis, $(y=0)$.

$U_1$: the y-axis, $(x=0)$.

$U_2$: the straight line $x=y$.

Then

$(1): V = W + U_1$, because coordinate wise sum of $(x,0)$ and $(0,y)$ will span whole of $R^2, (x,y)$,

$(2):$ any $v\in V$ can be uniquely represented by a unique element in $W$ and an unique element in $U_1$. $(x,y)=(x,0)+(0,y)$

So $V=W \bigoplus U_1$

Also,

$(1): V = W + U_2$, because coordinate wise sum of $(x,0)$ and $(y,y)$ will span whole of $R^2, (x,y)$,

$(2):$ any $v\in V$ can be uniquely represented by a unique element in $W$ and an unique element in $U_2$. $(x,y)=((x-y)+y,y)=((x-y),0)+(y,y)$

So $V=W \bigoplus U_2$

But, as we see, $U_1 \neq U_2$

My proof: $V=U_1\bigoplus W\Rightarrow \forall v\in V, \exists$ a unique $u\in U_1$ and $w\in W$, such that $v=u+w$.

Let us consider a specific $v_1=u_1+w_1......(1)$

$V=U_2\bigoplus W\Rightarrow \forall v\in V, \exists $ unique $u\in U_2, w\in W$, such that $v=u+w$.

With respect to the specific $v_1$ and $w_1$ in equation $(1), \exists$ a unique $u_2\in U_2$, satisfying $v_1=u_2+w_1.............(2)$

Comparing equations $(1)$ and $(2)$, we get $$u_1+w_1=u_2+w_1$$

Adding $(-1)w$ to both sides, we get $$u_1 + w_1 -w_1= u_2+w_1-w_1$$ $$\Rightarrow u_1+0 = u_2 +0$$ $$\Rightarrow u_1 = u_2.........(3)$$

Now, for any $u_1\in U_1, \exists $ unique $v_1, w_1$ and $u_2$, satisfying equation $(1)$ and $(2)$. And $u_1=u_2\in U_2$. In other words, $u_1\in U_1\Rightarrow u_1\in U_2$. Therefore $U_1\subseteq U_2.......(4)$

Similarly, for any $u_2\in U_2$, from equations $(1),(2) $ and $(3)$, we can show $u_2\in U_2 \Rightarrow u_2\in U_1$. Therefore $U_2\subseteq U_1.......(5)$.

Hence, from equations $(4)$ and $(5)$, we conclude $U_1=U_2$.

Answer 1

Comparing equations (1) and (2), we get u1+w1=u2+w1

There's no reason to say that the $w$ you get from $U_1 \oplus W$ is the same as the one you get from $U_2 \oplus W$.