[proofplan]
We construct $\tilde{T}$ by setting $\tilde{T}(x + Y) := T(x)$. The hypothesis $Y \subset \ker(T)$ is exactly the condition needed for this assignment to be independent of the coset representative. Linearity and uniqueness follow from the surjectivity of the quotient map $q: X \to X/Y$. For the norm equality, we use the fact (from the [Quotient Norm and the Quotient Map](/theorems/2631)) that $q$ maps the open unit ball of $X$ onto the open unit ball of $X/Y$, so $\tilde{T}$ and $T$ have the same image on unit balls, hence the same operator norm.
[/proofplan]
[step:Define $\tilde{T}$ on cosets and check well-definedness]
Define the candidate map
\begin{align*}
\tilde{T}: X/Y &\to Z \\
x + Y &\mapsto T(x).
\end{align*}
We verify that this assignment depends only on the coset $x + Y$, not on the chosen representative $x$. Suppose $x + Y = x' + Y$. Then $x - x' \in Y$. Since $Y \subset \ker(T)$ by hypothesis, we have $T(x - x') = 0$, so $T(x) = T(x')$ by linearity of $T$. Hence $\tilde{T}$ is well-defined as a map $X/Y \to Z$.
[guided]
We want to define $\tilde T$ so that $T = \tilde T \circ q$, where $q(x) = x + Y$ is the quotient map. The equation $T = \tilde T \circ q$ forces the value of $\tilde T$ at every coset: for $x + Y \in X/Y$, we must have $\tilde T(x + Y) = \tilde T(q(x)) = T(x)$. So there is no choice — the formula $\tilde T(x + Y) := T(x)$ is dictated by the factorisation requirement.
The only question is whether this prescription is **consistent**: a coset $x + Y$ has many representatives, and the formula assigns a value to each representative. For consistency, all representatives must yield the same output. If $x + Y = x' + Y$, then $x - x' \in Y$. We need $T(x) = T(x')$, i.e. $T(x - x') = 0$, i.e. $x - x' \in \ker(T)$. The hypothesis $Y \subset \ker(T)$ guarantees exactly this: $x - x' \in Y \subset \ker(T)$, so $T(x - x') = 0$, giving $T(x) = T(x')$.
This is precisely the universal property at work: the kernel hypothesis is the obstruction-removal step that lets the map descend to the quotient.
[/guided]
[/step]
[step:Verify linearity of $\tilde{T}$]
Let $x_1 + Y, x_2 + Y \in X/Y$ and $\lambda, \mu$ be scalars. By the definition of vector space operations on $X/Y$, we have $\lambda(x_1 + Y) + \mu(x_2 + Y) = (\lambda x_1 + \mu x_2) + Y$. Applying the definition of $\tilde T$ and using linearity of $T$:
\begin{align*}
\tilde{T}\bigl(\lambda(x_1 + Y) + \mu(x_2 + Y)\bigr) &= \tilde{T}\bigl((\lambda x_1 + \mu x_2) + Y\bigr) \\
&= T(\lambda x_1 + \mu x_2) \\
&= \lambda T(x_1) + \mu T(x_2) \\
&= \lambda \tilde{T}(x_1 + Y) + \mu \tilde{T}(x_2 + Y).
\end{align*}
Hence $\tilde T$ is linear.
[/step]
[step:Verify the factorisation identity $T = \tilde{T} \circ q$ and uniqueness]
For every $x \in X$,
\begin{align*}
(\tilde{T} \circ q)(x) = \tilde{T}(q(x)) = \tilde{T}(x + Y) = T(x),
\end{align*}
so $T = \tilde{T} \circ q$.
For uniqueness, suppose $S: X/Y \to Z$ is any linear map satisfying $T = S \circ q$. For an arbitrary element $x + Y \in X/Y$ we have, since $q(x) = x + Y$,
\begin{align*}
S(x + Y) = S(q(x)) = T(x) = \tilde{T}(x + Y).
\end{align*}
Because $q$ is surjective ([Quotient Norm and the Quotient Map](/theorems/2631)), every element of $X/Y$ has the form $x + Y$ for some $x \in X$, so $S = \tilde{T}$ on all of $X/Y$.
[/step]
[step:Show $\|\tilde{T}\| \le \|T\|$ via the quotient norm]
Let $x + Y \in X/Y$. By definition of the quotient norm, $\|x + Y\|_{X/Y} = \operatorname{dist}(x, Y) = \inf_{y \in Y} \|x + y\|_X$. For every $y \in Y$,
\begin{align*}
\|\tilde{T}(x + Y)\|_Z = \|T(x)\|_Z = \|T(x + y)\|_Z \le \|T\| \cdot \|x + y\|_X,
\end{align*}
where the second equality uses $T(y) = 0$ (because $y \in Y \subset \ker(T)$) and the inequality is the operator-norm bound for $T$. Taking the infimum over $y \in Y$ on the right-hand side:
\begin{align*}
\|\tilde{T}(x + Y)\|_Z \le \|T\| \cdot \inf_{y \in Y} \|x + y\|_X = \|T\| \cdot \|x + Y\|_{X/Y}.
\end{align*}
Hence $\tilde{T}$ is bounded with $\|\tilde{T}\| \le \|T\|$.
[guided]
We want to bound $\|\tilde T(x + Y)\|_Z$ by a constant multiple of $\|x + Y\|_{X/Y}$. The quotient norm is an infimum: $\|x + Y\|_{X/Y} = \inf_{y \in Y}\|x + y\|_X$. So bounding by the quotient norm means bounding by every $\|x + y\|_X$ — the bound has to remain valid as we choose better and better representatives.
The trick is that the value of $\tilde T(x + Y) = T(x)$ does not depend on the representative, but we are free to switch representatives when applying the bound for $T$. Indeed, for any $y \in Y$, the element $x + y$ also represents the coset $x + Y$, and since $y \in Y \subset \ker(T)$,
\begin{align*}
T(x) = T(x) + T(y) = T(x + y).
\end{align*}
This identity is what couples the kernel hypothesis to the quotient-norm bound: it lets us swap $x$ for any other representative $x + y$ before applying $\|T(\cdot)\| \le \|T\| \|\cdot\|$. We obtain
\begin{align*}
\|\tilde T(x + Y)\|_Z = \|T(x + y)\|_Z \le \|T\| \cdot \|x + y\|_X,
\end{align*}
and the bound holds for every $y \in Y$. Taking the infimum on the right gives $\|T\| \cdot \|x + Y\|_{X/Y}$, and the inequality $\|\tilde T(x + Y)\|_Z \le \|T\| \cdot \|x + Y\|_{X/Y}$ follows.
[/guided]
[/step]
[step:Show $\|\tilde{T}\| \ge \|T\|$ using the open-ball image of $q$]
By the [Quotient Norm and the Quotient Map](/theorems/2631), the quotient map $q$ maps the open unit ball $D_X = \{x \in X : \|x\|_X < 1\}$ surjectively onto $D_{X/Y} = \{\xi \in X/Y : \|\xi\|_{X/Y} < 1\}$. Combined with the factorisation $T = \tilde{T} \circ q$, this gives the equality of image sets
\begin{align*}
T(D_X) = \tilde{T}(q(D_X)) = \tilde{T}(D_{X/Y}).
\end{align*}
Taking the supremum of $\|\cdot\|_Z$ over each side:
\begin{align*}
\|T\| = \sup_{x \in D_X} \|T(x)\|_Z = \sup_{\xi \in D_{X/Y}} \|\tilde{T}(\xi)\|_Z = \|\tilde{T}\|.
\end{align*}
Combined with the upper bound from the previous step, $\|\tilde{T}\| = \|T\|$.
[guided]
The identity $T(D_X) = \tilde T(D_{X/Y})$ is a clean way to extract the norm equality without having to chase representatives. It rests on two facts already established:
1. **Factorisation:** $T = \tilde T \circ q$, so $T(D_X) = \tilde T(q(D_X))$ as sets.
2. **Open-ball surjectivity of $q$:** the [Quotient Norm and the Quotient Map](/theorems/2631) shows $q(D_X) = D_{X/Y}$ — the open unit ball of $X$ maps onto the open unit ball of $X/Y$. Concretely, for any $\xi = x + Y \in D_{X/Y}$, $\operatorname{dist}(x, Y) < 1$ means there is $y \in Y$ with $\|x + y\|_X < 1$, and then $x + y \in D_X$ and $q(x + y) = \xi$.
Combining these two facts: $T(D_X) = \tilde T(q(D_X)) = \tilde T(D_{X/Y})$. Now, the operator norm of any bounded linear map $A: V \to W$ between normed spaces equals $\sup_{v \in D_V}\|A(v)\|_W$ — this is the standard formula obtained by homogeneity. Applying it on both sides:
\begin{align*}
\|T\| = \sup_{x \in D_X}\|T(x)\|_Z = \sup_{\xi \in D_{X/Y}}\|\tilde T(\xi)\|_Z = \|\tilde T\|.
\end{align*}
Why does this argument need the **open** ball rather than just any neighbourhood? Because the supremum formula $\|A\| = \sup_{v \in D_V}\|A(v)\|_W$ is sharp for the open unit ball, and surjectivity of $q$ on this specific set is what is supplied by the quotient theorem.
[/guided]
[/step]
[step:Combine the bounds to conclude]
The previous two steps give $\|\tilde{T}\| \le \|T\|$ and $\|\tilde{T}\| \ge \|T\|$, hence $\|\tilde{T}\| = \|T\|$. Together with the well-definedness, linearity, and uniqueness established earlier, this completes the proof: $\tilde{T} \in \mathcal{L}(X/Y, Z)$ is the unique bounded linear map with $T = \tilde{T} \circ q$, and $\|\tilde{T}\| = \|T\|$.
[/step]
