# Morphism is a retraction of another morphism?

I have to use definition that morphism f is retraction of morphism g, but I don't know how to represent it. Any idea?

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Ah, this is related to your other question I just answered. The terminology isn't so good, because Hew's answer would be right most of the time, but in the context of model categories $f$ is a retraction of $g$ if it's a retract of $g$ in the category who objects are arrows in the original category $\mathcal{C}$ and whose morphisms are commutative squares. This is the diagram expressing the same information: $$\begin{matrix} A&\stackrel{i}{\to}&B&\stackrel{r}{\to}&A\\f\downarrow&&g\downarrow&&f\downarrow\\A'&\stackrel{j}{\longrightarrow}&B'&\stackrel{t}{\longrightarrow}&A'\end{matrix}$$ If $ri=1_A$ and $tj=1_{A'}$, we say $f$ is a retraction of $g.$

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"$f$ is a retraction of $g$" probably just means that $fg = 1$ in your category, whatever that category is. I don't know what you mean by "how to represent it". What are you trying to do?

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I'm using slice category, I'm not sure how diagram looks like when f is retraction of g. Actually, I have to prove that slice category is Quillen model category by checking axioms and my stuck with retract axiom. – dandan Oct 17 '12 at 20:38
Sorry, I'm not familiar with model categories. I'm sure someone around here is, though. I suggest you edit your question and its title to make it more specific, and maybe add a category-theory tag. Also, I see the Wikipedia article on model categories has a very specific diagram describing the retract axiom, so maybe that will help. – Hew Wolff Oct 17 '12 at 20:57
This is a comment, not an answer. – Graphth Oct 17 '12 at 21:54
I see your point, @Graphth. I thought dandan was having trouble applying the general definition of retraction, but now we know the story is more complicated. – Hew Wolff Oct 18 '12 at 3:20