If $V$ is a vector space and $U$ & $W$ are subspaces of $V$, such that $U \oplus W = V$! Need help with proofs! Consider the map $\rho : V \to V$, defined by $\rho(v) = u − w$, where $v = u + w$, $u \in U$, $w \in W$.
Show that:
i. $\rho$ is well defined and it is linear;
ii. $\rho(u) = u$, $\forall u ∈ U$;
iii. $\rho$ is an isomorphism and $ρ \circ ρ = 1V$ , where $1V$ denotes the identity map on $V$.
I have no idea how to do these, but just guessing these answers:
i. Because $\rho$ maps to itself it is well-defined, and it's linear because it's going from $V \to V$, and thus the dimension is singular and doesn't change?
ii. Because it goes from $V \to$, for all $u \in U$, there is a solution $\rho(u)$ that is in $V$?
iii. $1V$ denotes that it's identity is the <1> vector? It's an isomorphism, because it maps from $V\to V$ which is both the same co-domain and range, so they must be isomorphic because they're the exact same?
 A: Part i:
This question is entirely about using definitions.
First, we need to show that $\rho$ is well defined.  In other words, given a $v \in V$, there can only be one answer to $\rho(v)$.  In order to prove that $\rho$ is well defined, you need to use the definition of a direct sum (i.e. the $\oplus$ symbol).
Showing that $\rho$ is linear means showing that $\rho$ satisfies the definition of linearity.  The key thing to notice here is that if
$v_1 = u_1 + w_1$ and $v_2 = u_2 + w_2$, then 
$$
\rho(v_1 + \alpha v_2) = (u_1 + \alpha u_2) - (w_1 + \alpha w_2)
$$
Part ii:
It's enough to note that $u = u + 0$, and $0 \in W$.
Part iii:
In order to show that $\rho$ is an isomorphism, it suffices to show that its kernel (nullspace) is $\{0\}$.
For the next bit, show that for any $u$ and $w$, we have $\rho(\rho(u+w)) = u + w$.  Why is that enough?

More on well-definedness:
To prove that $\rho$ is well-defined, I would say the following:

For any $v$, there exist unique vectors $u \in U$ and $w \in W$ such that $v = u + w$.  Since there is exactly one choice of vectors $u,w$ corresponding to any $v$, the vector $\rho(v) = u - w$ is also uniquely determined.  So, $\rho$ is well-defined.

Consider, on the other hand, what this situation might look like if we didn't have $U \cap W = \{0\}$.  For example, take $U = \{(x,y,0):x,y \in \Bbb R\}$, and $W = \{(0,y,z):y,z \in \Bbb R\}$.
Take $v = (1,0,1)$. We might want to say that $\rho(1,0,1) = (1,0,-1)$ since we have $u = (1,0,0) \in U$ and $w = (0,0,1) \in W$ with $u + w = v$.  
However, we could also have taken $u = (1,1,0)$ and $w = (0,-1,1)$.  Because of this, we have the second answer $\rho(v) = u - w = (1,2,-1)$, which is clearly not the same as our first answer of $(1,0,-1)$.  So in this case, $\rho$ would not be well defined.
