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The slice category $\boldsymbol{C}/C$ of a cateogry $\boldsymbol{C}$ over an object $C\in\boldsymbol{C}$ has

  • Objects: All arrows $f\in \boldsymbol{C}$ such that $cod(f) = C$,
  • Arrows: an arrow $a$ from $f:X\to C$ to $f':X'\to C$ is an arrow $a:X\to X'$ in $\boldsymbol{C}$ such that $f'\circ a=f$

My question is how do we know that such an arrow as $a$ exists in $\boldsymbol{C}$? Or is this saying that if it exists, than we get arrows in slice category, and if it doesn't, our slice category can potentially be without any arrows?

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The only category without arrows is the empty category, since there's an identity arrow for each object. – joriki Oct 26 '12 at 23:22
up vote 3 down vote accepted

It's saying that the arrows in $\boldsymbol C/C$ from $f:X\to C$ (an arrow of $\boldsymbol C$ and an object of $\boldsymbol C/C$) to $f':X'\to C$ are precisely the arrows from $X$ to $X'$ in $\boldsymbol{C}$ such that $f'\circ a=f$. That's your second option, except it doesn't imply that the category can potentially be without any arrows, since for every $f:X\to C$ there is always the arrow from $f:X\to C$ to $f:X\to C$ given by the identity morphism on $X$.

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Oh I forgot about identity arrows. But ignoring those, I am glad my guess was correct, thanks! – drozzy Oct 26 '12 at 23:35
@joriki, In addition to the identity morphism on $X$ (from $f:X\to C$ to $f:X\to C$), is there not also necessarily an arrow connecting from $f:X\to C$ to $f:C\to C$ , because of composition? – smartcaveman Apr 24 '13 at 16:48
@joriki,… – smartcaveman Apr 24 '13 at 17:37

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