[proofplan]
We verify first that nonzero scalar multiplication defines an [equivalence relation](/page/Equivalence%20Relation) on $V_0 = V \setminus \{0\}$. Then we define the span map $F: V_0 \to \mathbb{P}(V)$ by sending a nonzero vector to the one-dimensional subspace it spans, and we prove that two nonzero vectors have the same image exactly when they are related by $\sim$. This fibre computation makes the induced quotient map $\Theta: V_0/{\sim} \to \mathbb{P}(V)$ well-defined and immediately proves both injectivity and surjectivity.
[/proofplan]
[step:Verify that nonzero scalar multiplication defines an equivalence relation]
Let $V_0 := V \setminus \{0\}$, and let $\sim$ be the relation on $V_0$ defined by
\begin{align*}
v \sim w \iff w = \lambda v \text{ for some } \lambda \in K^\times.
\end{align*}
We check the three equivalence-relation properties. For every $v \in V_0$, we have $v = 1_K v$ and $1_K \in K^\times$, so $v \sim v$.
If $v,w \in V_0$ and $v \sim w$, then there exists $\lambda \in K^\times$ such that $w = \lambda v$. Since $\lambda \in K^\times$, its inverse $\lambda^{-1}$ belongs to $K^\times$, and
\begin{align*}
v = \lambda^{-1} w.
\end{align*}
Hence $w \sim v$.
If $u,v,w \in V_0$, $u \sim v$, and $v \sim w$, then there exist $\lambda,\mu \in K^\times$ such that $v = \lambda u$ and $w = \mu v$. Substituting the first equality into the second gives
\begin{align*}
w = \mu \lambda u.
\end{align*}
Because $K^\times$ is closed under multiplication, $\mu\lambda \in K^\times$, so $u \sim w$. Thus $\sim$ is an equivalence relation.
[/step]
[step:Identify precisely when two nonzero vectors span the same projective point]
For $v \in V_0$, define
\begin{align*}
K v := \{\alpha v : \alpha \in K\}.
\end{align*}
Since $v \neq 0$, the subspace $K v$ is one-dimensional over $K$. Therefore the map
\begin{align*}
F: V_0 &\to \mathbb{P}(V)
\end{align*}
\begin{align*}
v &\mapsto K v
\end{align*}
is well-defined.
We prove that, for all $v,w \in V_0$,
\begin{align*}
F(v) = F(w) \iff v \sim w.
\end{align*}
Suppose first that $v \sim w$. Then $w = \lambda v$ for some $\lambda \in K^\times$. For every $\alpha \in K$, we have
\begin{align*}
\alpha w = \alpha \lambda v \in K v,
\end{align*}
so $K w \subset K v$. Also $v = \lambda^{-1} w$, and the same argument gives $K v \subset K w$. Hence $K v = K w$, so $F(v) = F(w)$.
Conversely, suppose $F(v) = F(w)$, so $K v = K w$. Since $w \in K w$ and $K w = K v$, there exists $\lambda \in K$ such that
\begin{align*}
w = \lambda v.
\end{align*}
Because $w \neq 0$, the scalar $\lambda$ cannot be $0$. Hence $\lambda \in K^\times$, and therefore $v \sim w$.
[guided]
The map to [projective space](/page/Projective%20Space) should send a nonzero vector to the line through the origin that it spans. Formally, for each $v \in V_0$, define
\begin{align*}
K v := \{\alpha v : \alpha \in K\}.
\end{align*}
This is a $K$-linear subspace of $V$. It is one-dimensional because $v \neq 0$ and the map $K \to K v$ given by $\alpha \mapsto \alpha v$ is a linear isomorphism onto its image. Thus the function
\begin{align*}
F: V_0 &\to \mathbb{P}(V)
\end{align*}
\begin{align*}
v &\mapsto K v
\end{align*}
is well-defined.
Now we determine exactly when two vectors have the same image under $F$. Let $v,w \in V_0$. If $v \sim w$, then by definition there is a scalar $\lambda \in K^\times$ such that $w = \lambda v$. Every element of $K w$ has the form $\alpha w$ for some $\alpha \in K$, and then
\begin{align*}
\alpha w = \alpha \lambda v \in K v.
\end{align*}
Therefore $K w \subset K v$. Since $\lambda$ is nonzero, $\lambda^{-1} \in K^\times$, and $v = \lambda^{-1}w$. Repeating the same containment argument with $v$ and $w$ interchanged gives $K v \subset K w$. Hence $K v = K w$, so $F(v) = F(w)$.
Conversely, suppose $F(v) = F(w)$. This means $K v = K w$. Since $w$ belongs to its own span $K w$, it also belongs to $K v$. By the definition of $K v$, there exists $\lambda \in K$ such that
\begin{align*}
w = \lambda v.
\end{align*}
The scalar $\lambda$ must be nonzero, because $\lambda = 0$ would give $w = 0$, contradicting $w \in V_0$. Therefore $\lambda \in K^\times$, and the defining condition for $v \sim w$ holds. We have proved
\begin{align*}
F(v) = F(w) \iff v \sim w.
\end{align*}
[/guided]
[/step]
[step:Descend the span map to the quotient and prove the induced map is bijective]
Define
\begin{align*}
\Theta: V_0/{\sim} &\to \mathbb{P}(V)
\end{align*}
\begin{align*}
[v]_{\sim} &\mapsto K v.
\end{align*}
This map is well-defined: if $[v]_{\sim} = [w]_{\sim}$, then $v \sim w$, so the previous step gives $K v = K w$.
The map $\Theta$ is injective. Indeed, if $\Theta([v]_{\sim}) = \Theta([w]_{\sim})$, then $K v = K w$. By the previous step, $v \sim w$, so $[v]_{\sim} = [w]_{\sim}$.
The map $\Theta$ is surjective. Let $L \in \mathbb{P}(V)$. By definition of $\mathbb{P}(V)$, the set $L$ is a one-dimensional $K$-linear subspace of $V$. Hence $L$ contains some vector $v \in V_0$, and since $L$ is one-dimensional and contains $v$, we have $L = K v$. Therefore
\begin{align*}
\Theta([v]_{\sim}) = L.
\end{align*}
Thus $\Theta$ is a well-defined bijection, which gives the claimed natural identification of $\mathbb{P}(V)$ with $(V \setminus \{0\})/{\sim}$.
[/step]
