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Let $\mathscr{C}$ be a model category (following Mark Hovey's "Model Categories") and let $*$ be its terminal object. Consider the slice category $*/\mathscr{C}$: it's endowed with a canonical forgetful functor $*/\mathscr{C} \to \mathscr{C}$ which has a left adjoint given by $X \mapsto X\coprod *$, with the inclusion in the coproduct as basepoint $* \to X\coprod *$. I can't understand why such a functor (the left adjoint) should be faithful, as claimed in Hovey's book at page 4. In fact this amounts to say that if two arrows $X\coprod * \to Y\coprod *$ of the form $f\coprod 1_*, g\coprod 1_*$ are equal, then $f=g$. This is of course true in the most common examples but why is it true in this generality?

Any hints? Thanks in advance.

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Since the functor is not faithful in general (see Zhen's answer), a natural question would be: Under what asssumptions is it faithful? –  Martin Brandenburg Dec 20 '13 at 10:50
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@MartinBrandenburg It suffices to assume that the coproduct insertion $Y \to Y \amalg 1$ is monic, which happens in e.g. any extensive category. –  Zhen Lin Dec 20 '13 at 11:57
    
Actually it's equivalent to that fact: I got stuck after realizing this, since I wasn't able to show that it holds. Moreover, it's enough to find a map $* \to Y$ for each $Y$, since in that case the coproduct injection would be a section, hence a mono. –  Lano Dec 20 '13 at 14:27
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1 Answer

The claim is false. Let $\mathcal{C}$ be the category of (commutative) rings. Then the terminal object in $\mathcal{C}$ is the trivial ring $\{ 0 \}$, and the slice under $\{ 0 \}$ is equivalent to the trivial category $\mathbb{1}$. In particular, the forgetful functor ${}^{\{ 0 \} /} \mathcal{C} \to \mathcal{C}$ cannot have a faithful left adjoint.

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Are you considering some trivial model structure on $\mathcal{C}$? Why should that be equivalent to the trivial category? It seems to me that is equivalente to $\mathcal{C}$ itself, but I can be wrong. –  Lano Dec 20 '13 at 10:10
    
The question has nothing to do with model categories. Also, in case it wasn't clear, my rings have unit and homomorphisms preserve them. So the only possible ring homomorphisms $\{ 0 \} \to R$ are those where $R$ is also trivial. –  Zhen Lin Dec 20 '13 at 10:12
    
Ok now the counterexample is clear; also in my opinion the model structure is useless here but then there must be a mistake in the book.. –  Lano Dec 20 '13 at 10:17
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Hovey's book is known to have mistakes. You can find errata here. –  Zhen Lin Dec 20 '13 at 10:21
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