[proofplan] We construct the map by sending the coset of $v$ modulo $\operatorname{Range}(i)$ to $q(v)$. Exactness at $V$ gives $\operatorname{Range}(i)=\ker(q)$, which is precisely the condition needed for this formula to be independent of the chosen representative. Linearity follows from the quotient operations and linearity of $q$, while exactness at $W$ gives surjectivity and exactness at $V$ gives injectivity. Finally, uniqueness follows because every element of the quotient is a coset of the form $v+\operatorname{Range}(i)$. [/proofplan] [step:Define the induced map on cosets and prove it is well-defined] Let \begin{align*} R:=\operatorname{Range}(i)\subset V. \end{align*} Since the sequence is exact at $V$, we have \begin{align*} R=\ker(q). \end{align*} Define a function \begin{align*} \Phi:V/R\to W \end{align*} by the rule \begin{align*} \Phi(v+R):=q(v) \end{align*} for each $v\in V$. We prove that this definition is independent of the representative. Suppose $v,v'\in V$ satisfy \begin{align*} v+R=v'+R. \end{align*} By equality of cosets, $v-v'\in R$. Since $R=\ker(q)$, this gives \begin{align*} q(v-v')=0. \end{align*} Using linearity of $q:V\to W$, \begin{align*} q(v)-q(v')=q(v-v')=0. \end{align*} Hence $q(v)=q(v')$, so $\Phi(v+R)$ is well-defined. [guided] Let \begin{align*} R:=\operatorname{Range}(i)\subset V. \end{align*} The reason for introducing $R$ is that the [quotient space](/page/Quotient%20Space) in the statement is $V/\operatorname{Range}(i)$, and the formula for $\Phi$ only depends on the coset modulo this subspace. Because the sequence is short exact, exactness at the middle space $V$ says that the image of the incoming map equals the kernel of the outgoing map: \begin{align*} R=\operatorname{Range}(i)=\ker(q). \end{align*} We now define a function \begin{align*} \Phi:V/R\to W \end{align*} by \begin{align*} \Phi(v+R):=q(v) \end{align*} for every $v\in V$. The only possible issue is that a coset has many representatives. Thus we must check that if $v+R$ and $v'+R$ are the same coset, then the formula gives the same value in $W$. Assume $v,v'\in V$ and \begin{align*} v+R=v'+R. \end{align*} By the definition of equality of cosets in a quotient [vector space](/page/Vector%20Space), this means \begin{align*} v-v'\in R. \end{align*} Since $R=\ker(q)$, we get \begin{align*} q(v-v')=0. \end{align*} Now use linearity of the map $q:V\to W$: \begin{align*} q(v)-q(v')=q(v-v')=0. \end{align*} Therefore $q(v)=q(v')$. This proves that the value assigned to a coset does not depend on the representative chosen, so $\Phi$ is well-defined as a map from $V/R$ to $W$. [/guided] [/step] [step:Use quotient operations to prove linearity] Let $\alpha\in k$ and let $v_1,v_2\in V$. By the definitions of addition and scalar multiplication in the quotient vector space $V/R$, \begin{align*} (v_1+R)+(v_2+R)=(v_1+v_2)+R \end{align*} and \begin{align*} \alpha(v_1+R)=(\alpha v_1)+R. \end{align*} Using the definition of $\Phi$ and linearity of $q$, we obtain \begin{align*} \Phi((v_1+R)+(v_2+R))=\Phi((v_1+v_2)+R)=q(v_1+v_2)=q(v_1)+q(v_2)=\Phi(v_1+R)+\Phi(v_2+R). \end{align*} Also, \begin{align*} \Phi(\alpha(v_1+R))=\Phi((\alpha v_1)+R)=q(\alpha v_1)=\alpha q(v_1)=\alpha\Phi(v_1+R). \end{align*} Hence $\Phi$ is $k$-linear. [/step] [step:Use exactness at $W$ to prove surjectivity] Let $w\in W$. Since the sequence is exact at $W$, the map $q:V\to W$ is surjective. Therefore there exists $v\in V$ such that \begin{align*} q(v)=w. \end{align*} For this $v$, the coset $v+R$ is an element of $V/R$, and by definition of $\Phi$, \begin{align*} \Phi(v+R)=q(v)=w. \end{align*} Thus $\Phi$ is surjective. [/step] [step:Use exactness at $V$ to prove injectivity] Let $v\in V$ and suppose \begin{align*} \Phi(v+R)=0. \end{align*} By definition of $\Phi$, this means \begin{align*} q(v)=0. \end{align*} Hence $v\in\ker(q)$. Since exactness at $V$ gives $\ker(q)=R$, we have $v\in R$. Therefore the coset $v+R$ is the zero element of $V/R$: \begin{align*} v+R=R. \end{align*} Thus the kernel of $\Phi$ is zero, so $\Phi$ is injective. [/step] [step:Conclude that the induced map is the unique isomorphism with the stated formula] We have proved that $\Phi:V/R\to W$ is linear, surjective, and injective. Therefore $\Phi$ is a linear isomorphism. It remains to prove uniqueness in the stated sense. Let \begin{align*} \Psi:V/R\to W \end{align*} be a [linear map](/page/Linear%20Map) such that \begin{align*} \Psi(v+R)=q(v) \end{align*} for every $v\in V$. Every element of $V/R$ has the form $v+R$ for some $v\in V$, and on each such element the value of $\Psi$ is forced to be $q(v)$, which is exactly $\Phi(v+R)$. Hence $\Psi=\Phi$. This proves the uniqueness of the linear map satisfying the displayed formula, and completes the proof. [/step]