[proofplan] We use the two defining properties of a complementary subspace: every vector of $V$ can be written as a sum of a vector in $U$ and a vector in $W$, and the only vector common to $U$ and $W$ is $0$. The map from $W$ to $V/U$ is the restriction of the quotient projection to $W$, so its linearity follows directly from the quotient [vector space](/page/Vector%20Space) operations. Surjectivity comes from representing each coset by the $W$-component of a decomposition in $V=U+W$, and injectivity comes from the condition $U\cap W=\{0\}$. [/proofplan] [step:Define the map from the complement to the quotient] Define \begin{align*} \iota:W\to V/U,\qquad \iota(w)=w+U. \end{align*} This is well-defined because every $w\in W$ is also an element of $V$, so the coset $w+U$ is an element of the quotient set $V/U$. [/step] [step:Verify that the map respects addition and scalar multiplication] Let $w_1,w_2\in W$ and let $a\in k$. Since $W$ is a linear subspace of $V$, we have $w_1+w_2\in W$ and $aw_1\in W$. Using the addition operation in $V/U$, \begin{align*} \iota(w_1+w_2)=(w_1+w_2)+U=(w_1+U)+(w_2+U)=\iota(w_1)+\iota(w_2). \end{align*} Using the scalar multiplication operation in $V/U$, \begin{align*} \iota(aw_1)=aw_1+U=a(w_1+U)=a\iota(w_1). \end{align*} Thus $\iota$ is linear. [guided] We need to prove that $\iota$ is a [linear map](/page/Linear%20Map) from $W$ to $V/U$. The first point is that the operations on $V/U$ are defined by representatives: \begin{align*} (v_1+U)+(v_2+U)=(v_1+v_2)+U \end{align*} and \begin{align*} a(v+U)=av+U. \end{align*} Now take arbitrary vectors $w_1,w_2\in W$ and a scalar $a\in k$. Because $W$ is a linear subspace, the vectors $w_1+w_2$ and $aw_1$ still lie in $W$, so both expressions $\iota(w_1+w_2)$ and $\iota(aw_1)$ are defined. For addition, applying the definition of $\iota$ and then the quotient addition rule gives \begin{align*} \iota(w_1+w_2)=(w_1+w_2)+U=(w_1+U)+(w_2+U)=\iota(w_1)+\iota(w_2). \end{align*} For scalar multiplication, applying the definition of $\iota$ and then the quotient scalar multiplication rule gives \begin{align*} \iota(aw_1)=aw_1+U=a(w_1+U)=a\iota(w_1). \end{align*} These two identities are exactly the additivity and homogeneity conditions for linearity, so $\iota:W\to V/U$ is linear. [/guided] [/step] [step:Use the decomposition $V=U+W$ to prove surjectivity] Let $x\in V/U$. By the definition of the quotient set, there exists $v\in V$ such that \begin{align*} x=v+U. \end{align*} Since $V=U+W$, choose $u\in U$ and $w\in W$ such that \begin{align*} v=u+w. \end{align*} Because $u\in U$, the cosets $(u+w)+U$ and $w+U$ are equal: indeed, \begin{align*} (u+w)-w=u\in U. \end{align*} Therefore \begin{align*} x=v+U=(u+w)+U=w+U=\iota(w). \end{align*} Thus every element of $V/U$ lies in the image of $\iota$, so $\iota$ is surjective. [/step] [step:Use the intersection condition to prove injectivity] Let $w\in W$ and suppose $\iota(w)=U$, where $U=0+U$ is the zero vector of $V/U$. Then \begin{align*} w+U=U. \end{align*} Equality of these cosets means $w-0\in U$, hence $w\in U$. Since also $w\in W$, we have \begin{align*} w\in U\cap W. \end{align*} By the hypothesis $U\cap W=\{0\}$, it follows that $w=0$. Therefore $\ker(\iota)=\{0\}$, and since $\iota$ is linear, $\iota$ is injective. [/step] [step:Conclude that the complement map is a linear isomorphism] We have proved that $\iota:W\to V/U$ is linear, surjective, and injective. Hence $\iota$ is a linear isomorphism. This is precisely the assertion that the complementary subspace $W$ identifies canonically with the [quotient space](/page/Quotient%20Space) $V/U$ through the map $w\mapsto w+U$. [/step]