---- All [[vector space|vector spaces]] are implicitly assumed over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. > [!proposition] Proposition. ([[characterization of quotienting a (semi)normed vector space]]) > The [[universal property|universal property]] [[characterization of quotienting a module|of]] [[quotient module|quotient vector spaces]] passes to both > 1. The [[subcategory]] $\mathsf{Seminorm_{T}}$ of [[seminorm|seminormed spaces]] [[continuous|and]] [[operator norm|bounded]] [[linear map|linear maps]]. > 2. The further [[subcategory]] $\mathsf{Seminorm}_{\mathsf{M}}$ of [[norm|seminormed spaces]] and [[Lipschitz continuous|nonexpansive]] [[linear map|linear maps]]. > > Explicitly: let $X,Y$ be [[norm|seminormed spaces]] and let $W \leq X$ be a [[linear subspace]]. We know that the [[universal property]] of [[quotient module|quotient vector spaces]] states that for any [[linear map|linear map]] $T:X\to Y$ satisfying $W \subset \operatorname{ker }T$,[^1] there exists a unique [[linear map]] making the following [[diagram]] commute: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAA0QBfU9TXfIRQBGclVqMWbAJrdeIDNjwEiZYePrNWiDgHoAOvog0YAJwZYwMYAHUucvksFEATGOqapOgBQAqAJTc4jBQAObwRKAAZqYQALZIZCA4EEiiElpsACoOIDHxidQpSC480bEJiG7JqYjpntoghjAAHlhwOHAABACEXYYA7liweAywwFn21Ax0AEYwDAAK-MpCIKZYoQAWOEFcQA > \begin{tikzcd} > X \arrow[r, "T"] \arrow[d] & Y & (*) \\ > X/{W} \arrow[ru, "\exists ! \widetilde{T}"'] & & > \end{tikzcd} > \end{document} > ``` > Here, $X/W$ is endowed with the [[quotient topology]].[^2] > > Then: > > **1.** *(Objects)* The [[quotient topology|(quotient)]] [[topological space|topology]] on $X / W$ is [[metric topology|induced]] by the [[seminorm]] $\|x + W\|:= \text{dist}(x, W)$ > witnessing $X/W$ to be a [[seminorm|seminormed space]]. (Equally, given $z \in X / W$, $\|z\|=\inf_{x \in \pi ^{-1} z} \|x\|$.) Also, $\pi(\mathbb{B}_{X}^{\circ})= \mathbb{B}_{X / Y}^{\circ}$. > > **2. ($\pi$ is in $\mathsf{T,M}$)** [[characterization of quotienting a topological space|By (the quotient topology) construction]], we know $\pi:X \to X /W$ is [[continuous]] (thus $\pi$ is in $\mathsf{Seminorm_{T}}$). This is also [[characterizing continuity of linear maps|witnessed by]] $\|\pi\|=1$ if $\overline{W} \neq X$ and $\|\pi\|=0$ [[dense|if]] $\overline{W}=X$ (thus $\pi$ is in fact in $\mathsf{Seminorm_{M}}$). > > **3.** *$(\mathsf{Seminorm_{T}})$-morphisms)* In general, $\|\widetilde{T}\| =\|T\|$. Hence if $T$ is [[operator norm|bounded]], then so is $\widetilde{T}$. > > **4.** *$(\mathsf{Seminorm_{M}})$-morphisms).* Following from (3), if $T$ is [[Lipschitz continuous|nonexpansive]] ($\|T\| \leq 1$) then so is $\widetilde{T}$. > > Put together, $(1)-(4)$ show that the [[diagram]] $(*)$, produced by the [[universal property]] of [[quotient set|quotients]] in the ambient [[category]] $\mathsf{Vect}$, in fact constitutes a valid [[diagram]] in $\mathsf{Seminorm_{T}}$ and $\mathsf{Seminorm_{M}}$. > > Lastly: in $\mathsf{Norm_{T}}$ and $\mathsf{Norm_{M}}$, replace $W$ by $\overline{W}$; then $X / \overline{W}$ is [[norm|normed]] and the discussion above goes through. > [!proposition] Corollary. (First bijection theorem for normed spaces) > The [[universal property]] of [[quotient module|quotient vector spaces]] entails that if $T:X \to Y$ is any [[linear map|linear map]] between [[seminorm|seminormed spaces]], then there is a unique [[linear map|linear map]] $\widetilde{T}$ making the diagram commute, and $\widetilde{T}$ is a [[linear isomorphism]] onto its image: > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAA0QBfU9TXfIRQBGclVqMWbAJrdeIDNjwEiZYePrNWiDgAIA9LuAAdYwGsYAJ10AVLnL5LBRUeuqapO0xDRW6OCEswOgBbGGAsEK5bBwV+ZSFkACYxd0ltEAAKACpsgEpucRgoAHN4IlAAM0sIEKQyEACkUQktNhtY6tr66ibEFJAACxg6KDYcAHcIYdGEHiqauv7eiCQAZmoZsZ1J6ZGoBF66LAY2QYgIMxBqBiwwDKg6OGGxtLavYxgADyw4HDhdABCXSmCZYWB4BiwYB2a4gBh0ABGMAYAAV4s4dJYsCVBjhOot1itmkcTmcLlcuBQuEA > \begin{tikzcd} > X \arrow[r, "T"] \arrow[d, two heads] & Y & (**) \\ > X / {\ker T} \arrow[r, "\exists ! \widetilde{T}"', two heads, dashed, hook] & \operatorname{im} T \arrow[u, hook] & > \end{tikzcd} > \end{document} > ``` > > > Augmenting with the universal properties above for $\mathsf{Seminorm_{T}}$ resp. $\mathsf{Seminorm_{M}}$ automatically produces the following "[[first isomorphism theorem|first bijection theorem]]s" for $\mathsf{Seminorm_{T}}$ resp. $\mathsf{Seminorm_{M}}$. > > **1. (First bijection theorem for $\mathsf{Seminorm_{T}}$)** If $T$ is a [[operator norm|bounded linear map]], then so is $\widetilde{T}$ in the [[diagram]] $(* *)$. > > **2. (First bijection theorem for $\mathsf{Seminorm_{M}}$)** In fact, $\|T\|=\|\widetilde{T}\|$, so that if $T$ is [[Lipschitz continuous|nonexpansive]] then so is $\widetilde{T}$ in the [[diagram]] $(* *)$. > > > > > Now, when can we upgrade to a true [[first isomorphism theorem|first]] *[[isomorphism]]* [[first isomorphism theorem|theorem]] rather than merely a "first *bijection* theorem"? That is, when is $\widetilde{T}^{-1}: \operatorname{im }T \to X / \operatorname{ker }T$ also a morphism in $\mathsf{Seminorm_{T}}$ resp. $\mathsf{Seminorm_{M}}$? (We call $T$ a **quotient operator (onto its image)**[^9] in such a case.) This is characterized below. > > > > [!proposition] Characterization of quotient operators. (For what $T$ does FIT hold?) > > > > For $T:X \to (\operatorname{im }T \subset Y)$ a [[operator norm|bounded linear map]] between [[seminorm|seminormed spaces]], the following are equivalent: > > 1. $\widetilde{T}^{-1}$ is bounded (i.e. $T$ is a quotient operator onto its image) > > 2. $\widetilde{T}$ is an [[open map]] ($\iff$ $T$ is an [[open map]] onto its image) > > 3. There exists $M>0$ such that for all $y \in \operatorname{im }T$ there exists $x \in X$ with $Tx=y$ and $\|x\| \leq M \| y\|$[^8] > > 4. $T(\mathbb{B})$ [[absorbing set|absorbs]] a neighborhood of $0$ in $\operatorname{im }T$ (that is, contains some ball around $0$ in $\operatorname{im } T$) > > 5. $T(\mathbb{B})$ has nonempty interior in $\operatorname{im }T$. > >> > > > Moreover, $T$ is an [[Lipschitz continuous|isometric]] quotient operator onto its image if and only if $T(\mathbb{B}_{X})=\mathbb{B}_{Y}$ . [^8]: This is a weaker condition than saying $T$ is bounded below, as we require the bound only for one representative per fiber. [^9]: "Onto its image" because quotient operators are required to be surjective. > [!proof] > > ^proof $(4)$ follows from $(3)$; so just have to show $(1)$-$(3)$. **1.** We have to show that the [[distance from point to set]] indeed defines a [[seminorm]] on $X / W$. Then we have to show this induces the [[quotient topology]]. - [ ] TODO **2.** If $\overline{W}=X$, then $\|\pi(x)\|=\text{dist}(x, W)=0$ for all $x,w$, hence $\|\pi\|=0$. Assume $\overline{W}\neq X$. Suppose $\overline{W} \neq X$. Then if $x \in X$ with $\|x\|=1$, have $\|\pi(x)\|=\inf_{w \in \overline{W}} \|x-w\| \leq \|x\|$ by looking at $w=0$. So $\|\pi\| \leq 1$. To see $\|\pi\| \geq 1$, obtain $x \in X-\overline{W}$. [[supremum|From properties]] of [[infimum]] there exists a [[monotonic map|monotone]]-decreasing [[sequence]] $\|x-w_{n} \|$ in $\{ \|x-w\|: w \in W \}$ converging to the infimum $\text{dist}(x,W)=\|x+W\|$. Put $y_{n}:=\frac{x-w_{n}}{\|x-w_{n}\|}$. Then $\|y_{n}\|=1$ and since $w_{n}\in W$, $\|\pi y_{n}\|=\frac{\text{dist}(x, W)}{\|x-w_{n}\|} \to \frac{\text{dist}(x, W)}{\text{dist}(x, W)}=1.$ It follows that $\|y\|=\sup_{\|y\|=1}\|\pi y\| \geq 1$. (If a sequence bounded above by some supremum (here $\|\pi\|=\sup_{\|y\|=1} \|\pi y\|$) converges to a limit $L$, then the supremum is at least $L$) Hence we've shown $\|\pi\| \geq 1$. As for $\pi(\mathbb{B}_{X}^{\circ})=\mathbb{B}_{X/Z}^{\circ}$, one inclusion follows just from $\|\pi\| \leq 1$. For the reverse inclusion, recall $\|x+Z\| = \inf_{x' \in \pi ^{-1} (x+Z)} \|x'\|$. Thus having $\|x+Z\| \leq 1$ implies there exists $x'$ with $\|x'\| \leq 1$ and $\pi(x')=x+Z$. **3.** By abstract considerations, $\|T\|=\|\widetilde{T} \circ \pi\|\leq \|\widetilde{T}\| \overbrace{ \|\pi\| }^{ =1 }$. For the other inequality $\geq$, let $x+W \in \frac{X}{W}$ such that $\|x+W\|=\operatorname{dist}(x,W)=1$. Have $\|\widetilde{T}(x+W)\|=\|Tx\|=\|T(x-w)\| \leq \|T\| \|x-w\|.$ Taking the infimum over $w \in W$ gives $\|\widetilde{T}(x+W)\|\leq \|T\| \underbrace{ \text{dist}(x,W) }_{ \|x+W\| }$. Hence $\|\widetilde{T}\| \leq \|T\|$. Just take $x':=x$; then there is equality by definition of $\widetilde{T}$. $\|\pi\|=1$, that $\|\widetilde{T}\|=\|T\|$ is immediate from the fact that, $\{ x+W \in X / W : \|x + W\|=1 \}$ $\{ x \in X : \|x\| =1 \}$. by definition, $\|\widetilde{T}(x+W)\|=\|Tx\|$, and passing to [[supremum|supremums]] on both sides. $\leq.$ Let $x+W \in \frac{X}{W}$ such that $\|x+W\|=\operatorname{dist}(x,W)=1$. Want to find $x \in X$, $\|x\|=1$, such that $\|\widetilde{T}(x+W)\| \leq \|Tx\|$. $\|\widetilde{T}(x+W)\|=\|Tx\|$ - [ ] TODO **4.** (characterization of quotient operators) **4.1.** By definition. **4.2.** $(1) \iff (2)$ Since the canonical projection and inclusion are open maps, $\widetilde{T}$ is open map iff $T$ is open map. **4.3.** $(1) \iff (3)$ If $T$ is a quotient operator, then $\widetilde{T} ^{-1}: \operatorname{im } T \to \frac{X}{\operatorname{ker } T}$ is a bounded and this means precisely that there exists $r>0$ such that $\|\underbrace{ \widetilde{T}x }_{ y }\| \geq r \|x\|$ for all $x$... **4.4.** $(4) \implies (1)$ Fix $r>0$ with $r\mathbb{B}_{Y}^{\circ} \subset T(\mathbb{B}_{X}^{\circ})$. WLOG $T$ surjective. Claim there exists $c>0$ such that $\|\widetilde{T} ^{-1} y\| \leq c \|y\|$ for all $y \in Y$. Clear if $y=0$. So let $y \in Y$ nonzero. Normalize to $\frac{y}{\|y\|} \in \mathbb{B}_{Y}$; by assumption $r_{0} \frac{y}{\|y\|} \in T(\mathbb{B}_{X}^{\circ})$ for some $0<r_{0}<r$ that is fixed for all $y$, e.g. $r_{0}=\frac{r}{2}$. (We can't just say $r \frac{y}{\|y\|}$ because it might be a boundary point, so must nudge into the interior.) Thus there exists $z \in X$, $\|z\| < 1$, with $\frac{r_{0}y}{\|y\|}=Tz$. Thus $\begin{align} \|\widetilde{T} ^{-1} y \| & = r_{0}^{-1} \|y\| \, \| \widetilde{T} ^{-1} (Tz) \| \\ &= r_{0}^{-1} \|y\| \, \underbrace{ \| \pi (z)\| }_{ < 1 } \\ & < r_{0} ^{-1}\| y\|. \end{align}$ where we've used $\| \pi\| \leq 1$. It then holds for $T$ also because the canonical projection is known to map the open ball onto the open ball, as is the canonical inclusion. **4.5.** $(4)\implies (5)$ is trivial. We show $(5)\implies (2)$. Assume $T\mathbb{B}_{X}$ has nonempty [[topological interior|interior]], as witnessed by some ball $\mathbb{B}_{Y}^{\circ}(Tx_{0}, r)$, $x_{0} \in \mathbb{B}_{X}^{\circ}, r>0$ inside. Note that, by [[balanced set|symmetry]], also $\mathbb{B}_{Y}^{\circ}(-Tx_{0}, r) \subset T\mathbb{B}_{X}$, and then by [[convex set|convexity]], $\underbrace{ \frac{1}{2} \mathbb{B}_{Y}^{\circ}(-Tx_{0}, r) + \frac{1}{2} \mathbb{B}_{Y}^{\circ}(Tx_{0}, r) }_{ = \mathbb{B}_{Y}^{\circ}(r, 0) } \subset T\mathbb{B}_{X}.$ So WLOG the witnessing ball is centered at $0$ of radius $r$. The claim is that $T$ is an open map. Let $U$ be open in $X$. Claim $T(U)$ is open; let $y \in T(U)$. Obtain $x \in U$ with $Tx=y$. Then since $U$ is open there exist $x_{} \in X$, $s>0$ such that $x_{}+s\mathbb{B}_{X}^{} \subset U$. Now $y \in T(x_{}+s \mathbb{B}_{X}^{})=Tx_{}+s T\mathbb{B}_{X} \subset T(U)$, and since $T\mathbb{B}_{X}$ contains $\mathbb{B}_{Y}^{\circ}(r)$ we have $y \in Tx_{}+ s \mathbb{B}_{Y}^{\circ }(r) \subset T(U)$ and this shows $T(U)$ is open. ---- #### [^1]: We could equivalently [[quotient set|quotient]] by an [[equivalence relation]] $\sim$ (as is [[quotient space|common in]] [[topological space|topology]]) and place the condition $(x \sim y) \implies (Tx=Ty)$ on $T$ ([[universal property of quotient sets|cf.]]). The discussion in [[characterization of quotienting a group|quotient group]] carries over to show that the collection of equivalence relations on $X$ equals the collection of subspaces we could mod out by. [^2]: As induced by the [[equivalence relation]] $v_{1} \sim v_{2} \iff (v_{1}-v_{2}) \in \operatorname{ker }(\pi:V \to V / W)$, ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```