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1. • CommentRowNumber1.
   • CommentAuthorzskoda
   • CommentTimeAug 20th 2024
   • (edited Aug 20th 2024)
   • PermaLink
   Author: zskoda
   Format: MarkdownItexMore generally, in any category $C$, a monomorphism $i_U : U\hookrightarrow X$, and a [[morphism]] $f \colon X\to Y$, the _[[restriction]]_ $f|_U \colon U \to Y$ of $f$ onto $U$ is the [[composition|precomposition]] $f|_U \coloneqq f \circ i_U$ of $f$ by $i_U$. A [[subobject]] is an equivalence class of monomorphisms. For a different representative of the subobject, $i_{\tilde{U}} \colon \tilde{U}\to X$ there is a unique [[isomorphism]] $b \colon U\to\tilde{U}$ such that $i_{\tilde{U}}\circ b = i_U$, hence $f_{\tilde{U}} = f\circ b$. <a href="https://ncatlab.org/nlab/revision/diff/restriction/4">diff</a>, <a href="https://ncatlab.org/nlab/revision/restriction/4">v4</a>, <a href="https://ncatlab.org/nlab/show/restriction">current</a>

   More generally, in any category $C$, a monomorphism $i_U : U\hookrightarrow X$, and a morphism $f \colon X\to Y$, the restriction $f|_U \colon U \to Y$ of $f$ onto $U$ is the precomposition $f|_U \coloneqq f \circ i_U$ of $f$ by $i_U$. A subobject is an equivalence class of monomorphisms. For a different representative of the subobject, $i_{\tilde{U}} \colon \tilde{U}\to X$ there is a unique isomorphism $b \colon U\to\tilde{U}$ such that $i_{\tilde{U}}\circ b = i_U$, hence $f_{\tilde{U}} = f\circ b$.

   diff, v4, current