# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMay 18th 2018

fixed the statement of Example 5.2 (this example) by restricting it to $\mathcal{C} = sSet$

• CommentRowNumber2.
• CommentAuthorYuxi Liu
• CommentTimeJul 2nd 2020

I’m just here to check the code since there’s no way to check in the editor.

Proof: Take any isomorphism $f$, let $f = gh$ and $h f^{-1} = g' h'$ be the unique factorizations. Then $id = ghf^{−1} = (gg')h'$, so $h' = id$ and $gg' = id$, whence $g = id$ and $g' = id$ since $g, g' \in R_+$. Thus $f = h \in R_−$. The same argument applied to $f^{−1}$ shows that $f$ preserves the degree, hence $f = id$.

• CommentRowNumber3.
• CommentAuthorYuxi Liu
• CommentTimeJul 2nd 2020

I’m just here to check the code since there’s no way to check in the editor.

Proof: Take any isomorphism $f$, let $f = gh$ and $h f^{-1} = g' h'$ be the unique factorizations. Then $id = gh f^{−1} = (gg')h'$, so $h' = id$ and $gg' = id$, whence $g = id$ and $g' = id$ since $g, g' \in R_+$. Thus $f = h \in R_−$. The same argument applied to $f^{−1}$ shows that $f$ preserves the degree, hence $f = id$.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeJul 2nd 2020

For checking code for the $n$Lab best to use the Sandbox page.

The parser here on the $n$Forum does not behave identically to that for $n$Lab pages.

• CommentRowNumber5.
• CommentAuthorDmitri Pavlov
• CommentTimeJan 31st 2021

Redirect: latching map, matching map.