Is Friedberg's proof wrong? I'm now reviewing Linear Algebra. As the picture, Theorem 1.3 said that $W$ is a subspace of $V$ if and only if


*

*(a) $0_V\in W$

*(b) $\forall x,~y\in W,~x+y\in W$

*(c) $\forall c\in F,~\forall x\in W,~cx\in W$.


In the "Conversely" part of the proof, if (a), (b), (c) holds, the author want to show that $W$ is a subspace of $V$. However, even if (a), namely $0_V\in W$ holds, we don't know that $0_V$ is the zero vector of $W$. He didn't show this. And this, is not so trivial that can be just said without a proof.
And next, the author went to prove that every element in $W$ has an inverse. He said that, by (c), we know that $(-1)v\in W$, and on the other hand, $(-1)v=-v\in W$. Here arises a critical question, what does this $-v$ corresponding to? Is it: 


*

*corresponding to $V$. i.e., $v+(-v)=0_V=(-v)+v$?

*or corresponding to $W$. i.e., admit that $0_W$ exists, and at the same time $v+(-v)=0_W=(-v)+v$?


By the context, it should be corresponding to $V$, namely $-v_V$. And then, the author must have to show that this $-v_V$, is the inverse element with respect to $W$. And this, again, is not so trivial that can even be stated without some arguments. I think Friedberg mistaken the role of $0$ and $-v$ here, leading he omitted some key part in the proof. Do you agree? If the proof of this theorem is a part of a mid-term exam for undergraduate students, and if a student write the proof just in that way, I think there would be few points. 

Updated: (the page before this theorem)

 A: Well, it is obvious that $0_V$ must be the  zero vector of $W$. It satisfies $0+v=v$ for all elements in $v\in V$, and in particular those in $W$. There can only be one element with that property, of course.
The same can be said about the additive inverse of any element of $W$. If you agree that $v$ has an additive inverse "in $W$", then you must immediately see why it is the same as the one "in $V$."
Everything seems totally clear as written, IMO.
A: In the "conversely" part of the proof, we assume conditions (a), (b), and (c) and have to show that $W$ is a vector space with the addition and scalar multiplication inherited from $V$. 
To prove $W$ is a vector space, we have to show that there is a vector $z \in W$ that is an additive identity for vector addition. By condition (a), $0_V$ is in $W$, and by the "prior discussion" this is a zero vector for $W$.  In the "conversely" part we are not assuming that $W$ is a vector space, and so there is not a pre-existing $0_W$ that we could refer to.  
Moreover, as the author points out just before the theorem statement, it follows from the first part of the proof that the only option for the zero vector of a subspace $W$ is the zero vector of $V$. So, in the second part, when we are trying to prove that $W$ is a subspace, we had better take the zero vector of $W$ to be the zero vector of $V$!
Similarly, to show that each vector in $W$ has an additive inverse in $W$, if $u \in W$ then $-u \in V$ and so $-u = (-1)w$ is in $W$ by condition (c).  Also $u + -u = 0$ because this was already true in $V$ and $W$ has the same addition for vectors in $W$.
Because we do not yet know $W$ is a vector space, we can't assume that there is a pre-existing additive inverse operation in $W$. Instead, we prove that the restriction of the additive inverse relation on $V$ is an additive inverse relation on $W$. 
