1
$\begingroup$

*The slice category or over category $C/c$ of a category $ C $ over an object $ c∈C $ has

objects that are all arrows $ f∈C $ such that $ cod(f)=c $,

and

morphisms $ g:X→X'∈C $ from $ f:X→c $ to $ f':X'→c $ such that $ f'∘g=f $.

*There is a forgetful functor $ U_{c}: C/c→C $ which maps an object $ f: X→c$ to its domain $ X $ and a morphism $ g:X→X'∈C/c $ (from $f:X→c$ to $f':X'→c$ such that $f'∘g=f$) to the morphism $g: X→X'$.

Please i need to know how one can define free objects using the free functor in that category.

$\endgroup$
3
  • $\begingroup$ It is not very clear what you mean by free objects in this context (you are not working over a concrete category). Do you mean to ask whether your forgetful functor has a left adjoint ? $\endgroup$ Commented Nov 13, 2019 at 20:57
  • $\begingroup$ Yes that what i actually mean $\endgroup$
    – Imane
    Commented Nov 13, 2019 at 21:06
  • $\begingroup$ Then in general it won't exist. Try to work out a simple example, like $C$ is the category of sets, and $c=\{0,1\}$. $\endgroup$ Commented Nov 13, 2019 at 21:21

1 Answer 1

3
$\begingroup$

The forgetful functor does not have a left adjoint in general, because the identity on $c$ is the terminal object of $C/c$, even if $U_c(id_c)=c$ is not a terminal object in $C$. In fact, if $c$ is terminal then $U_c$ is an isomorphism, so $U_c$ has a left adjoint if and only if it has an inverse.

On the other hand, if $C$ has products then $U_c$ has a right adjoint, which takes an object $X$ to the projection $X\times c\to c$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .