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draft/draft.tex

@@ -2281,7 +2281,7 @@ Now, we want to abstract the given proof for an extensive category that has term
 		\item $h=Fg\comp k$: In this case we take $k'=k$. Now, $k'\appr k$ and $g+1\comp k'=h$.\qed
 	\end{itemize}
 \end{proof}
-To have an abstraction of~\autoref{lem:maybe-func-set} recalling that our category extensive, we use the fact that for an arbitrary morphism $f\c X\to Y+Z$, we can have morphisms $f_Y\c X_Y\to Y$ and $f_Z\c X_Z\to Z$, such that $X_Y+X_Z\iso X$, so we follow with $f_Y+f_Z$. In our case, we take $\brks{\alpha\comp p_1,\alpha\comp p_2}\c R\to (X+1)\times(X+1)$, where $(X+1)\times (X+1)\iso X^2+(2\times X)+1$, and we assume the following pullbacks exist:
+To have an abstraction of~\autoref{lem:maybe-func-set} recalling that our category is extensive, we use the fact that for an arbitrary morphism $f\c X\to Y+Z$, we can have morphisms $f_Y\c X_Y\to Y$ and $f_Z\c X_Z\to Z$, such that $X_Y+X_Z\iso X$, so we follow with $f_Y+f_Z$. In our case, we take $\brks{\alpha\comp p_1,\alpha\comp p_2}\c R\to (X+1)\times(X+1)$, where $(X+1)\times (X+1)\iso X^2+(2\times X)+1$, and we assume the following pullbacks exist:
 \begin{equation*}
 	\begin{tikzcd}[ampersand replacement=\&]
 		{R_{X^2}} \& R \& {R_{2\times X}} \& R \\