Prove that $\eta: V \to V/W$ defined as $\eta(x)=x+W$ is a surjective linear transformation, and find $\ker(\eta)$. Prove that $\eta: V \to V/W$ defined as $\eta(x)=x+W$ is a surjective linear transformation, and $\ker(\eta)=W$.
My proof:
Let $y \in V/W$.  Then let $x=y+W$. So $\eta(x)=\eta(y+W)=(y+W)+W=y+W$
I don't think that last step is right at all.  I'm not too sure how to find the kernel either.
Thanks in advance.
 A: The elements of $V/W$ are the sets of the form $x+W$ for $x\in V$, so if $y\in V/W$, then $y$ already is $x+W$ for some $x\in V$. That means that $y=\eta(x)$, so we’ve just proved that $\eta$ is surjective. You still have to prove that $\eta$ is linear and find $\ker\eta$.
To show that $\eta$ is linear, you have to show that for every $x,y\in V$ and scalars $\alpha,\beta$, $$\eta(\alpha x+\beta y)=\alpha\eta(x)+\beta\eta(y)\;.$$ To do that, use the definition of $\eta$ to write down what $\eta(\alpha x+\beta y)$ and $\alpha\eta(x)+\beta\eta(y)$ are, and see if you can see (and explain!) why they must be equal.
The kernel of $\eta$ is the set of vectors $x\in V$ such that $\eta(x)$ is the zero vector in $V/W$. What is that zero vector? Once you figure out what it is, it’s not to hard to figure out which elements $x+W$ of $V/W$ are equal to it.
A: Hint:  What is the zero vector of $V/W$?
A: Your proof needs slight modification. Let $y \in V/ W$. By definition of what it means to be an element of $V / W$ we may choose $x \in V$ such that $y = x + W$. Then $\eta(x) = x+W = y$ which shows $\eta$ is surjective.
Then we need to show that $\ker(\eta) = W$. Firstly, we must consider what the zero vector in $V / W$ is. It is the equivalence class of $0\in V$. Hence, it is $W$. Let any $x \in W$ be given, then we certainly have $\eta(x) = x+W=W$ so $\ker(\eta) \supseteq W$. Alternatively, suppose $x \in V$ but $x \notin W$, then $\eta(x) = x+W \neq W$, so $x \notin \ker(\eta)$.
It follows that $\ker(\eta) = W$.
Note: In the previous proof we used the fact that $x +W = W \iff x \in W$ since $W$ is a subspace.
