In set theory, suppose $A\subset B$. Then suppose we want to "replace $A$ by $C$". We have a notation for this: $B'=(B\setminus A) \cup C$.

However, suppose that $A,B,C$ are groups, or topological spaces, and $A$ is a subgroup or subspace of $B$, and that we want to generate a new group or topological space $B'$ by replacing $A$ by $C$. In this case, in order to remain faithful to the structure, we have to specify a particular way in which it's replaced, so we need extra data. Suppose we have a morphism $\phi:C\to A$ (homomorphism or continuous function), which tells us how the replacement is supposed to happen.

Is there a standard notation for denoting the equivalent of replacement $B'=(B\setminus A) \cup C$, in general categories other than Set? (I am not sure actually which assumptions about the category would be needed to define this).

Edit: I can imagine that something like $B[\phi:A\setminus C]$ is standard notation?

  • $\begingroup$ What do you mean by "replacing"? what purpose does it serve? $\endgroup$ – Jackozee Hakkiuz Aug 1 at 13:09
  • 2
    $\begingroup$ Before discussing notation, you should first define your construction in detail. Think about the topological space case first – that should be comparatively straightforward. After that you might think about the group case. When you have a construction that works in a few well known categories, come back and ask your question again. $\endgroup$ – Zhen Lin Aug 1 at 13:10

You may be looking for Heyting arrow. It sometimes happens that the poset of subobjects of a fixed object $A$ is also a Heyting algebra. This means that it's a distributive lattice and that for every pair of subobjects $B$ and $C$, there is a subobject denoted $B\implies C$ or $B\succ C$ such that $$ (D\wedge B) \leq C \hspace{5mm} \text{if and only if} \hspace{5mm} D\leq(B\succ C).$$

In the case of subsets of a set $A$, then $\wedge$ is intersection, $\leq$ is containment and I claim that $B\succ C = B^c\cup C$, where $B^c$ is the absolute complement (i.e., with respect to $A$). You can check this: if $D\cap B\subseteq C$, then $D\subseteq C$ and hence $D\subseteq B^c\cup C$. Conversely, if $D\subseteq B^c\cup C$, then $D\cap B \subseteq (B^c\cup C)\cap B=C$.

This also happens if you are considering the family of open subsets of a topological space. In this case you have $B\succ C = \text{Int}(B^c\cup C)$.

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