[proofplan] Define the candidate map on a coset by sending $v+U$ to $T(v)$. The only point requiring care is well-definedness: if two vectors represent the same coset, their difference lies in $U$, and the hypothesis $U\subset\ker(T)$ forces $T$ to take the same value on both representatives. Once the map is well-defined, linearity follows directly from the quotient [vector space](/page/Vector%20Space) operations, the identity $T=\widetilde{T}\circ q$ follows by evaluating at each vector of $V$, and uniqueness follows because every element of $V/U$ is a coset $q(v)$. [/proofplan] [step:Define the candidate map on cosets and prove it is well-defined] Define a map \begin{align*} \widetilde{T}:V/U\to W \end{align*} as follows. For an element $v+U\in V/U$, set \begin{align*} \widetilde{T}(v+U):=T(v). \end{align*} We verify that this definition does not depend on the representative $v$. Let $v,v'\in V$ and suppose \begin{align*} v+U=v'+U. \end{align*} By equality of cosets in the [quotient space](/page/Quotient%20Space), this is equivalent to \begin{align*} v-v'\in U. \end{align*} Since $U\subset\ker(T)$, we have $T(v-v')=0_W$, where $0_W$ denotes the zero vector of $W$. Because $T$ is linear, \begin{align*} T(v-v')=T(v)-T(v'). \end{align*} Therefore \begin{align*} T(v)-T(v')=0_W, \end{align*} so $T(v)=T(v')$. Hence $\widetilde{T}(v+U)$ is independent of the chosen representative, and $\widetilde{T}:V/U\to W$ is a well-defined map. [guided] We want to define the induced map by the natural formula \begin{align*} \widetilde{T}(v+U):=T(v). \end{align*} The possible problem is that an element of $V/U$ is not a vector $v\in V$, but a coset $v+U$. The same coset can have many representatives, so we must prove that choosing a different representative gives the same value in $W$. Let $v,v'\in V$ be two representatives of the same coset, so \begin{align*} v+U=v'+U. \end{align*} By the definition of equality of cosets in the quotient space, this means exactly that \begin{align*} v-v'\in U. \end{align*} The hypothesis $U\subset\ker(T)$ is designed precisely for this point: since $v-v'\in U$, we get \begin{align*} v-v'\in\ker(T). \end{align*} By the definition of the kernel, this says \begin{align*} T(v-v')=0_W. \end{align*} Using the linearity of $T:V\to W$, we compute \begin{align*} T(v-v')=T(v)-T(v'). \end{align*} Combining the last two displayed identities gives \begin{align*} T(v)-T(v')=0_W. \end{align*} Thus \begin{align*} T(v)=T(v'). \end{align*} So the value assigned by the formula $\widetilde{T}(v+U)=T(v)$ depends only on the coset $v+U$, not on the chosen representative $v$. Therefore the formula defines a genuine map \begin{align*} \widetilde{T}:V/U\to W. \end{align*} [/guided] [/step] [step:Verify that the induced map is linear] Let $v_1,v_2\in V$ and let $a,b\in k$. By the vector space operations on $V/U$, we have \begin{align*} a(v_1+U)+b(v_2+U)=(av_1+bv_2)+U. \end{align*} Using the definition of $\widetilde{T}$ and the linearity of $T$, we obtain \begin{align*} \widetilde{T}\bigl(a(v_1+U)+b(v_2+U)\bigr)=\widetilde{T}((av_1+bv_2)+U). \end{align*} Hence \begin{align*} \widetilde{T}\bigl(a(v_1+U)+b(v_2+U)\bigr)=T(av_1+bv_2). \end{align*} Since $T$ is linear, \begin{align*} T(av_1+bv_2)=aT(v_1)+bT(v_2). \end{align*} By the definition of $\widetilde{T}$ again, \begin{align*} aT(v_1)+bT(v_2)=a\widetilde{T}(v_1+U)+b\widetilde{T}(v_2+U). \end{align*} Therefore \begin{align*} \widetilde{T}\bigl(a(v_1+U)+b(v_2+U)\bigr)=a\widetilde{T}(v_1+U)+b\widetilde{T}(v_2+U). \end{align*} Thus $\widetilde{T}$ is linear. [/step] [step:Check that the factorization identity holds] We prove that \begin{align*} T=\widetilde{T}\circ q. \end{align*} Let $v\in V$. Since $q(v)=v+U$, the definition of $\widetilde{T}$ gives \begin{align*} (\widetilde{T}\circ q)(v)=\widetilde{T}(q(v)). \end{align*} Thus \begin{align*} (\widetilde{T}\circ q)(v)=\widetilde{T}(v+U). \end{align*} By the defining formula for $\widetilde{T}$, \begin{align*} \widetilde{T}(v+U)=T(v). \end{align*} Hence \begin{align*} (\widetilde{T}\circ q)(v)=T(v) \end{align*} for every $v\in V$. Therefore $T=\widetilde{T}\circ q$. [/step] [step:Prove uniqueness from surjectivity of the quotient map] Let \begin{align*} S:V/U\to W \end{align*} be a [linear map](/page/Linear%20Map) such that \begin{align*} T=S\circ q. \end{align*} We prove that $S=\widetilde{T}$. Let $x\in V/U$. By the definition of the quotient space, there exists $v\in V$ such that \begin{align*} x=v+U. \end{align*} Equivalently, $x=q(v)$. Therefore \begin{align*} S(x)=S(q(v)). \end{align*} Since $S\circ q=T$, we have \begin{align*} S(q(v))=T(v). \end{align*} By the definition of $\widetilde{T}$, \begin{align*} T(v)=\widetilde{T}(v+U). \end{align*} Since $x=v+U$, this gives \begin{align*} S(x)=\widetilde{T}(x). \end{align*} Because $x\in V/U$ was arbitrary, $S=\widetilde{T}$. Thus the linear map $\widetilde{T}:V/U\to W$ satisfying $T=\widetilde{T}\circ q$ exists and is unique, with formula $\widetilde{T}(v+U)=T(v)$ for every $v\in V$. [/step]