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- \documentclass{article}
- \usepackage{amsmath}
- \usepackage{amsthm}
- \usepackage{amssymb}
- \usepackage{amsfonts}
- \usepackage{url}
- \usepackage{xspace}
- \usepackage{xy}
- \usepackage{comment}
- \xyoption{all} \CompileMatrices
- \newcommand{\Zobr}[3]{{#1}\colon{#2}\to{#3}}
- \newcommand{\ol}[1]{\overline{#1}}
- %\addtolength{\hoffset}{-1cm} \addtolength{\textwidth}{2cm}
- %\includecomment{mine}
- \excludecomment{mine}
- \begin{document}
- The fact that $P$ is [projective]%(http://en.wikipedia.org/wiki/Projective_object)
- means the following:
- Whenever we have an epimorphism $\Zobr fXY$ and a morphism $\Zobr pPY$, then there exists a morphism $\Zobr{\ol p}PX$ such that $f\circ\ol p=p$.
- $$
- \xymatrix@C=10pt@R=17pt{
- & {P} \ar@{-->}[ld]_{\overline{p}} \ar[rd]^{p} \\
- {X} \ar[rr]_{f} && {Y} }
- $$
- Now we have the following situation
- $$
- \xymatrix@C=10pt@R=17pt{
- & P \ar@<-3pt>[d]_{r} \\
- & A \ar[rd]^{p} \ar@<-3pt>[u]_{s} \\
- X \ar[rr]_f && Y
- }
- $$
- The rest of solution is: draw the obvious arrows to complete the diagram.
- This would be a logical place to stop, if you only want a hint and want to do the rest by yourself.
- Since it is already some time since you posted your question, I guess posting full solution will not do much harm.
- (And of course you can simply ignore the rest of you post.)
- We have the morphism $q=p\circ r$. Since $P$ is projective, there is a morphism $\ol q$ such that $f\circ \ol q=q=p\circ r$. Thus we get
- $$f\circ\ol q \circ s=p\circ r\circ s=p\circ 1_A=p,$$
- i.e. we have shown that there is a morphism $\ol p = \ol q\circ s$ fulfilling $f\circ\ol p=p$.
- $$
- \xymatrix@C=10pt@R=17pt{
- & P \ar@<-3pt>[d]_{r} \ar[rdd]^q \ar[ldd]_{\overline q} \\
- & A \ar[rd]_{p} \ar@<-3pt>[u]_{s} \ar[ld]^{\overline p} \\
- X \ar[rr]_f && Y
- }
- $$
- \end{document}
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