Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For an object $c$ of a site $C$ with terminal object $*$ there is a functor \begin{equation} f:\operatorname{Sh}(C)/c\to \operatorname{Sh}(C)/*=\operatorname{Sh}(C) \end{equation} from the slice topos into the whole topos given by composiong $F\to c$ with $c\to *$ where I abuse the notation for the Yoneda embedding.

What is an example of such a situation such that $f$ does not preserve monomorphisms?

share|cite|improve this question
@ShawnHenry: No, this functor is a left adjoint, and cannot be a right adjoint in general because it does not preserve products. – Zhen Lin Oct 25 '12 at 22:47
up vote 1 down vote accepted

There is no such situation, and the reason is purely category-theoretic.

Let $\mathcal{C}$ be any category. I claim that the monomorphisms in the slice category $(\mathcal{C} \downarrow Z)$ are exactly the same as the monomorphisms in $\mathcal{C}$. It's not hard to see that if a morphism in $(\mathcal{C} \downarrow Z)$ is a monomorphism in $\mathcal{C}$ then it must also be a monomorphism in $(\mathcal{C} \downarrow Z)$ because composition in $(\mathcal{C} \downarrow Z)$ is inherited from $\mathcal{C}$. On the other hand, suppose we have a monomorphism $f : A \to B$ in $(\mathcal{C} \downarrow Z)$, and two arrows $g, h : X \to A$ in $\mathcal{C}$ such that $f \circ g = f \circ h$. Then, if we denote by $a : A \to Z$ and $b : B \to Z$ the structural morphisms, we must have $$a \circ g = b \circ f \circ g = b \circ f \circ h = a \circ h$$ and so we can think of $g, h : X \to A$ as morphisms in $(\mathcal{C} \downarrow Z)$ as well, and thus $f \circ g = f \circ h$ implies $g = h$, as required to be a monomorphism in $\mathcal{C}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.