双線型空間の普遍性
$ \forall K\in{\cal B}\forall U,V\in{\cal L}(K)に対し、$ T\in{\cal L}(K)と「普遍的な双線型写像」$ b:\underline U\times\underline V\to\underline Tが存在する $ bが普遍的であることは、以下と同値である
$ \forall W\in{\cal L}(K)\forall f:\underline U\times\underline V\to\underline W\exists!\bar f:\underline T\to\underline W.f=\bar f\circ b\land\bar f\text{は線型写像である}
https://scrapbox.io/files/65c6ea2fd411da00259aa3e9.svg
$ bにより、$ \underline U\times\underline V\to\underline Wから$ \underline T\to\underline Wへの一対一対応を構成できる
$ \forall K\in{\cal B}\forall U,V\in{\cal L}(K)\forall b:\underline U\times\underline V\to\underline T\forall b':\underline U\times\underline V\to\underline T'.b,b'\text{が普遍性を満たす}\implies T\cong T'
ここで
$ T\cong T':\iff\exist\phi:T\to T'.\phi\text{は同型写像である}
証明
$ \forall K\in{\cal B}\forall U,V\in{\cal L}(K)\forall b:\underline U\times\underline V\to\underline T\forall b':\underline U\times\underline V\to\underline T'にて
$ b,b'\text{が普遍性を満たす}
$ \iff\forall W\in{\cal L}(K).
https://scrapbox.io/files/65c6ee651d64320024a9b569.svg
$ \implies
https://scrapbox.io/files/65c6ee7814f8450024d79665.svg
左図式の$ Wに$ T'を、右図式の$ Wに$ Tを代入した
あとで使うので$ bの普遍性は残してある
$ \iff\forall W\in{\cal L}(K)\exists! \bar f,\bar f'\text{ are linear}.
https://scrapbox.io/files/65c6ee9a73a5a20025f2a291.svg
図式を合体しただけ
記号がかぶらないよう、右図式の$ \bar fを$ \bar f''に変えた
$ \implies
https://scrapbox.io/files/65c6eeb33244aa0024786647.svg
左図式の上矢印を合成し、真ん中の$ b'を削った
右に恒等写像$ {\rm id}_T:T\to Tの図式を加えた これは恒等式なので常に成り立つ
$ \implies\exists! \bar f,\bar f'\text{ are linear}.\bar f'\circ\bar f={\rm id}_T
ここが証明の核心
真ん中の図式即ち$ bの普遍性が示す一意性より、左図式の点線上の写像$ \bar f'\circ fと右図式の点線上の写像$ {\rm id}_Tが等しくなる
$ \implies \exist f:T\to T'.f\text{は同型写像である}
さっきも書いたが、同型写像は全単射と等しいはず
$ \iff T\cong T'
$ \underline{\therefore\forall K\in{\cal B}\forall U,V\in{\cal L}(K)\forall b:\underline U\times\underline V\to\underline T\forall b':\underline U\times\underline V\to\underline T'.b,b'\text{が普遍性を満たす}\implies T\cong T'\quad}_\blacksquare
この定理から、どのようにテンソル積を定義しても、実質同じだとわかる
テンソル積を、基底を使わずに定義できたということ
数カ月ぶりにこのページに書き込んだが、当時はこの定理の意味を全然わかってなかったのだと思う
わかってなかったということがわかるのも成長か
code:eg0.6.tikz(tex)
\begin{document}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T$};
\node (W) at (0,2) {$\underbar W$};
\draw (S) -- node {$b$} (V);
\draw (S) -- nodeswap {$\forall f$} (W); \drawdashed (V) -- node {$\exists! \bar{f}\textrm{ is linear}$} (W); \end{tikzpicture}
\end{document}
code:lemma0.7.tikz(tex)
\begin{document}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T$};
\node (W) at (0,2) {$\underbar W$};
\draw (S) -- node {$b$} (V);
\draw (S) -- nodeswap {$\forall f$} (W); \drawdashed (V) -- node {$\exists! \bar{f}\textrm{ is linear}$} (W); \end{tikzpicture}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T'$};
\node (W) at (0,2) {$\underbar W$};
\draw (S) -- node {$b'$} (V);
\draw (S) -- nodeswap {$\forall f$} (W); \drawdashed (V) -- node {$\exists! \bar{f}\textrm{ is linear}$} (W); \end{tikzpicture}
\end{document}
code:lemma0.7-2.tikz(tex)
\begin{document}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T$};
\node (W) at (0,2) {$\underbar T'$};
\draw (S) -- node {$b$} (V);
\draw (S) -- nodeswap {$b'$} (W); \drawdashed (V) -- node {$\exists! \bar{f}\textrm{ is linear}$} (W); \end{tikzpicture}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T'$};
\node (W) at (0,2) {$\underbar T$};
\draw (S) -- node {$b'$} (V);
\draw (S) -- nodeswap {$b$} (W); \drawdashed (V) -- node {$\exists! \bar{f}\textrm{ is linear}$} (W); \end{tikzpicture}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T$};
\node (W) at (0,2) {$\underbar W$};
\draw (S) -- node {$b$} (V);
\draw (S) -- nodeswap {$\forall f$} (W); \drawdashed (V) -- node {$\exists! \bar{f}\textrm{ is linear}$} (W); \end{tikzpicture}
\(\textrm{.for }\forall W\in{\cal L}(K)\)
\end{document}
code:lemma0.7-3.tikz(tex)
\begin{document}
\node (UV) at (2,0) {$\underbar U\times\underbar V$};
\node (T) at (0,-2) {$\underbar T$};
\node (T') at (0,0) {$\underbar T'$};
\node (T2) at (0,2) {$\underbar T$};
\draw (UV) -- node {$b$} (T);
\draw (UV) -- nodeswap {$b$} (T2); \draw (UV) -- nodeswap {$b'$} (T'); \drawdashed (T) -- node {$\bar f$} (T'); \drawdashed (T') -- node {$\bar f'$} (T2); \end{tikzpicture}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T$};
\node (W) at (0,2) {$\underbar W$};
\draw (S) -- node {$b$} (V);
\draw (S) -- nodeswap {$\forall f$} (W); \drawdashed (V) -- node {$\exists! \bar f''\textrm{ is linear}$} (W); \end{tikzpicture}
\end{document}
code:lemma0.7-4.tikz(tex)
\begin{document}
\node (UV) at (2,0) {$\underbar U\times\underbar V$};
\node (T) at (0,0) {$\underbar T$};
\node (T2) at (0,2) {$\underbar T$};
\draw (UV) -- node {$b$} (T);
\draw (UV) -- nodeswap {$b$} (T2); \drawdashed (T) -- node {$\bar f'\circ\bar f$} (T2); \end{tikzpicture}
\node (S) at (2,0) {$\underbar U\times\underbar V$};
\node (V) at (0,0) {$\underbar T$};
\node (W) at (0,2) {$\underbar W$};
\draw (S) -- node {$b$} (V);
\draw (S) -- nodeswap {$\forall f$} (W); \drawdashed (V) -- node {$\exists! \bar f''\textrm{ is linear}$} (W); \end{tikzpicture}
\node (UV) at (2,0) {$\underbar U\times\underbar V$};
\node (T) at (0,0) {$\underbar T$};
\node (T2) at (0,2) {$\underbar T$};
\draw (UV) -- node {$b$} (T);
\draw (UV) -- nodeswap {$b$} (T2); \drawdashed (T) -- node {${\rm id}_T$} (T2); \end{tikzpicture}
\end{document}