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Let $\sf C$ be a category with all small limits and colimits. Let $*$ be the terminal object, and denote by $\sf C_*$ the under category $*/\sf C$.

Define the functor $I:\sf C\to C_*$ by setting $I(c)$ as the canonical morphism $*\to (*\sqcup c) $ into the coproduct. I read in Hovey's Model Categories that, in case $\sf C$ has a model structure, $I$ is an embedding, i.e. it is (1) faithful and (2) injective on objects.

(1) Take two morphisms $f,g:c'\to c$ in $\sf C$. Let $i_c,i_*$ be the canonical morphisms into $*\sqcup c$; then the components of $I(f)$ are $(i_c\circ f, i_*)$, and similarly those of $I(g)$ are $(i_c\circ g, i_*)$. Hence $I_{c',c}$ is injective iff $i_c$ is monic and, in order to prove that $I$ is faithful, I'd prove that the canonical morphism $c\to (*\sqcup c)$ is monic for any object $c$. I don't have a clue on how to use the model structure to get this done.

(2) Let $c,c'$ be objects such that $* \sqcup c=* \sqcup c'$, in a way that the canonical morphisms $*\to * \sqcup c$ and $*\to * \sqcup c'$ are equal too. I shall prove that $c=c'$, which seems quite strong; are you sure that I shouldn't prove just $c\cong c'$? Anyway, I can't even obtain a morphism $c\to c'$, let alone an isomorphism.

I'm new to model categories but I have a background in category theory, and this problem was at the beginning of Hovey's book (like at page 4). So I guess these problems should be easy checks for the reader, while I don't even know how to start them; is this because I should read other texts before Hovey, or should these checks be easy for anyone practical with category theory? Thanks for any help.

I add below the paragraph of the book (the last of page 4) which I quoted, to be sure that it's not a misunderstanding of mine.

enter image description here

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    $\begingroup$ The claim about faithfulness is a mistake. The existing errata for this book (people.math.rochester.edu/faculty/doug/otherpapers/…) was last updated on January 6, 1999, so it is not surprising that this mistake is not mentioned there. $\endgroup$
    – Dmitri P.
    Commented Apr 7, 2023 at 16:29
  • $\begingroup$ @DmitriP. also the claim that it is an embedding, i.e. injective on objects, is false, right? $\endgroup$ Commented Apr 7, 2023 at 16:35
  • $\begingroup$ The terminology “embedding of categories” has multiple inequivalent definitions. Any functor can be replaced by an equivalent functor that is injective on objects, and any functor can be replaced by an equivalent functor that is not injective on objects. But the answer to the question as it is stated is affirmative, there are examples where the functor is not injective on objects. $\endgroup$
    – Dmitri P.
    Commented Apr 7, 2023 at 16:40
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    $\begingroup$ Even if we were to be generous and interpret it as "injective on isomorphism classes of objects" or "conservative", $\textbf{CRing}$ is a counterexample. For that matter, any category with a strict terminal object would be a counterexample. $\endgroup$
    – Zhen Lin
    Commented Apr 7, 2023 at 17:06
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    $\begingroup$ “But the only functor S→S′, equivalent to f, is f itself”: this is not true. Present S→S' as the composition S→S⊔S'→S', where the first map is the canonical inclusion and the second map is induced by f and the identity map on S'. Create a unique isomorphism in S⊔S' for every pair of objects that maps to the same element of S'. Now the functor S⊔S'→S' is an equivalence of categories. Therefore, the functor f is equivalent to the functor S→S⊔S', which is injective on objects. $\endgroup$
    – Dmitri P.
    Commented Apr 7, 2023 at 17:55

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The statement isn't true. Any category admits a model structure, so if the statement were true then $\mathcal{C} \to \mathcal{C}_{*}$ would be faithful and injective on objects for any complete and cocomplete category $\mathcal{C}$. But this isn't true. For example, if $\mathcal{C}$ is the category of commutative rings then $\mathcal{C}_{*}$ is the terminal category, and the unique functor $\mathcal{C} \to \mathcal{C}_{*}$ is neither faithful nor injective on objects.

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