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1. • CommentRowNumber1.
   • CommentAuthorSven Keidel
   • CommentTimeJul 17th 2016
   • (edited Jul 17th 2016)
   • PermaLink
   Author: Sven Keidel
   Format: MarkdownItexHello, I am looking for a notion of filtering the arrows of a category, if they can be composed with the arrows of another category, that is: Let $\mathbb{C}$ and $\mathbb{D}$ be categories with the same objects. The result of the filtering should be a category with the arrows of $\mathbb{C}$, that post-compose with arrows of $\mathbb{D}$: $$\lbrace g \in \mathbb{D}_1 \mid \exists f \in \mathbb{C}_1. \mathit{cod}(f) = \mathit{dom}(g) \rbrace$$ Is there a way I can state this clearly with category theory instead of relying on this set builder notation? Thanks, Sven

   Hello, I am looking for a notion of filtering the arrows of a category, if they can be composed with the arrows of another category, that is:

   Let $\mathbb{C}$ and $\mathbb{D}$ be categories with the same objects. The result of the filtering should be a category with the arrows of $\mathbb{C}$, that post-compose with arrows of $\mathbb{D}$:

   $$\lbrace g \in \mathbb{D}_1 \mid \exists f \in \mathbb{C}_1. \mathit{cod}(f) = \mathit{dom}(g) \rbrace$$

   Is there a way I can state this clearly with category theory instead of relying on this set builder notation?

   Thanks, Sven

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJul 19th 2016
   • PermaLink
   Author: Urs
   Format: MarkdownItexSince there exist identity morphisms, every object in a category is the source and the domain of some morphism. So the condition in your set builder is empty and the set you write down is just that of morphisms of $\mathbb{D}$.

   Since there exist identity morphisms, every object in a category is the source and the domain of some morphism. So the condition in your set builder is empty and the set you write down is just that of morphisms of $\mathbb{D}$.

3. • CommentRowNumber3.
   • CommentAuthorSven Keidel
   • CommentTimeJul 19th 2016
   • (edited Jul 19th 2016)
   • PermaLink
   Author: Sven Keidel
   Format: MarkdownItexOk, I understood my problem better. I can rephrase my question to make it more clear: Given two functions $f: A \rightarrow B$, $g: B' \rightarrow C$, construct a function $f_{\star}g: f^{-1}(B \cap B') \rightarrow C$ with $f_{\star}g(a) = g(f(a))$, where $f^{-1}(B\cap B') = \lbrace a \in A \mid f(a) \in B \cap B' \rbrace$. I.e. the situation looks like this: $$ \array{ f^{-1}(B \cap B') & \xrightarrow{\overline{f}} & B \cap B' & \xrightarrow{j'} & B' & \xrightarrow{g} & C \\ \downarrow & & \downarrow^{\mathrlap{j}} & & & & \\ A &\xrightarrow{f} & B & & & & } $$ So the candidate for $f_{\star}g$ would be $g \circ j' \circ \overline{f}$.

   Ok, I understood my problem better. I can rephrase my question to make it more clear:

   Given two functions $f: A \rightarrow B$, $g: B' \rightarrow C$, construct a function $f_{\star}g: f^{-1}(B \cap B') \rightarrow C$ with $f_{\star}g(a) = g(f(a))$, where $f^{-1}(B\cap B') = \lbrace a \in A \mid f(a) \in B \cap B' \rbrace$. I.e. the situation looks like this:

   $$\array{ f^{-1}(B \cap B') & \xrightarrow{\overline{f}} & B \cap B' & \xrightarrow{j'} & B' & \xrightarrow{g} & C \\ \downarrow & & \downarrow^{\mathrlap{j}} & & & & \\ A &\xrightarrow{f} & B & & & & }$$

   So the candidate for $f_{\star}g$ would be $g \circ j' \circ \overline{f}$.

4. • CommentRowNumber4.
   • CommentAuthorSven Keidel
   • CommentTimeJul 19th 2016
   • PermaLink
   Author: Sven Keidel
   Format: MarkdownItexFor categories other than $\mathbf{Set}$, we would have to require that $B \cap B'$ and $f^{-1}(B \cap B')$ are pullbacks and $j$ and $j'$ are monomorphisms.

   For categories other than $\mathbf{Set}$, we would have to require that $B \cap B'$ and $f^{-1}(B \cap B')$ are pullbacks and $j$ and $j'$ are monomorphisms.

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeJul 19th 2016
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   Author: DavidRoberts
   Format: MarkdownItexWhat do you mean by $B\cap B'$? This makes no sense outside of a material set theory context. You need some maps $B \to Z \leftarrow B'$ to get a pullback, and they need to be monomorphisms to get anything like what you are suggesting. At the moment, I still can't see what you're getting at, and not sure where the conversation is going.

   What do you mean by $B\cap B'$? This makes no sense outside of a material set theory context. You need some maps $B \to Z \leftarrow B'$ to get a pullback, and they need to be monomorphisms to get anything like what you are suggesting.

   At the moment, I still can’t see what you’re getting at, and not sure where the conversation is going.

6. • CommentRowNumber6.
   • CommentAuthorSven Keidel
   • CommentTimeJul 19th 2016
   • (edited Jul 19th 2016)
   • PermaLink
   Author: Sven Keidel
   Format: MarkdownItex@DavidRoberts, I formulated the problem in terms of sets and functions and now I try to generalize it to $\mathbf{Cat}$. You are right, $B \cap B'$ makes no sense in outside of set theory, but how can I find and object, that acts like $B \cap B'$ in $\mathbf{Cat}$? In $\mathbf{Set}$ an object $B \cap B'$ with $B,B' \subseteq Z$ satisfies the following property: $$ \array{ B \cap B' & \xrightarrow{i'} & B' \\ \downarrow^{\mathrlap{i}} & & \downarrow^{\mathrlap{j'}} \\ B & \xrightarrow{j} & Z & } $$ Where $i,i',j,j'$ are subset inclusion maps and $B\cap B'$ is the pullback of Z along $j$ and $j'$.

   @DavidRoberts, I formulated the problem in terms of sets and functions and now I try to generalize it to $\mathbf{Cat}$.

   You are right, $B \cap B'$ makes no sense in outside of set theory, but how can I find and object, that acts like $B \cap B'$ in $\mathbf{Cat}$? In $\mathbf{Set}$ an object $B \cap B'$ with $B,B' \subseteq Z$ satisfies the following property:

   $$\array{ B \cap B' & \xrightarrow{i'} & B' \\ \downarrow^{\mathrlap{i}} & & \downarrow^{\mathrlap{j'}} \\ B & \xrightarrow{j} & Z & }$$

   Where $i,i',j,j'$ are subset inclusion maps and $B\cap B'$ is the pullback of Z along $j$ and $j'$.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJul 19th 2016
   • (edited Jul 19th 2016)
   • PermaLink
   Author: Urs
   Format: MarkdownItexYou had already said it correctly in #4: an [[intersection]] of sets is a special kind of [[fiber product]] of sets (hence of [[pullback]] of sets), namely one where both input morphisms are monomorphisms. Hence if the diagram in #3 is what you would like to generalize to categories other than $Set$ then it's straightforward: for $f \colon A \to B$ and $g \colon B' \to C$ two morphism in your category, and if $B \to Z \leftarrow B'$ are understood, then, assuming that the respective two pullback squares (pb) exist in your category (for instance if it has all finite limits), then you seem to want to consider the diagram $$ \array{ A \underset{Z}{\times} B' &\longrightarrow& B \underset{Z}{\times} B' &\longrightarrow& B' &\overset{g}{\longrightarrow}& C \\ \downarrow &(pb)& \downarrow &(pb)& \downarrow \\ A &\underset{f}{\longrightarrow}& B &\longrightarrow& Z } \,. $$ Here I am using the [[pasting law]] law to identify the object in the top left as shown.

   You had already said it correctly in #4: an intersection of sets is a special kind of fiber product of sets (hence of pullback of sets), namely one where both input morphisms are monomorphisms.

   Hence if the diagram in #3 is what you would like to generalize to categories other than $Set$ then it’s straightforward:

   for $f \colon A \to B$ and $g \colon B' \to C$ two morphism in your category, and if $B \to Z \leftarrow B'$ are understood, then, assuming that the respective two pullback squares (pb) exist in your category (for instance if it has all finite limits), then you seem to want to consider the diagram

   $$\array{ A \underset{Z}{\times} B' &\longrightarrow& B \underset{Z}{\times} B' &\longrightarrow& B' &\overset{g}{\longrightarrow}& C \\ \downarrow &(pb)& \downarrow &(pb)& \downarrow \\ A &\underset{f}{\longrightarrow}& B &\longrightarrow& Z } \,.$$

   Here I am using the pasting law law to identify the object in the top left as shown.

8. • CommentRowNumber8.
   • CommentAuthorSven Keidel
   • CommentTimeJul 19th 2016
   • PermaLink
   Author: Sven Keidel
   Format: MarkdownItex@Urs, @DavidRoberts, thanks, this was really helpful. Keep up the good work on ncatlab.

   @Urs, @DavidRoberts, thanks, this was really helpful. Keep up the good work on ncatlab.