Timeline for Why this morphism of stacks is an isomorphism?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 31 at 17:53 | comment | added | Tian Vlašić | What @Kevin Carlson is trying to say is that when you write $f: X \rightarrow Y$, it means that $X$ (that is, $Y$) is the domain (that is, the codomain) of the morphism $f$. This has nothing to do with whether or not $f$ is a function. But, if $f$ is a function, then we write $f: x \mapsto y_x$ to mean that $f(x)=y_x$ for all $x \in X$. It is misleading to use $\mapsto$ for $\rightarrow$. | |
| Jun 3 at 11:19 | comment | added | Analyse300 | Indeed! You are right all morphism are not mappings! Thanks. | |
| Jun 3 at 6:43 | comment | added | Kevin Carlson | You are using the wrong arrow $\mapsto$ where you should write $\to.$ The $\mapsto$ arrow is used to describe a mapping on terms, as in $f:\mathbb R\to \mathbb R, x\mapsto x^2.$ | |
| Jun 2 at 16:38 | comment | added | Analyse300 | It works! Many thanks! | |
| Jun 2 at 13:25 | comment | added | R. van Dobben de Bruyn | This is Tag 046N, isn't it? | |
| Jun 2 at 8:23 | history | asked | Analyse300 | CC BY-SA 4.0 |