For a partial function $p:X\to Y$, we have

  • $p^{-1}(A\cap B)=p^{-1}(A)\cap p^{-1}(B)$
  • $p^{-1}(A\cup B)=p^{-1}(A)\cup p^{-1}(B)$
  • $p^{-1}(A \setminus B)=p^{-1}(A) \setminus p^{-1}(B)$
  • $p^{-1}(A \operatorname{\Delta} B)=p^{-1}(A) \operatorname{\Delta} p^{-1}(B)$ where $A \operatorname{\Delta} B := (A \setminus B) \cup (B \setminus A)$

But $p^{-1}(Y)=X$ is only true if $p$ is a total function. Especially $p^{-1}(A^c)=p^{-1}(A)^c$ is only true (even for specific $A$) if $p$ is total, since otherwise $$p^{-1}(Y)=p^{-1}(A\cup A^c)=p^{-1}(A)\cup p^{-1}(A^c) \quad = \quad p^{-1}(A)\cup p^{-1}(A)^c=X$$

We also have

  • $A \subseteq B \Rightarrow p^{-1}(A) \subseteq p^{-1}(B)$
  • $A' \subseteq B' \Rightarrow p(A') \subseteq p(B')$
  • $p(p^{-1}(B)) \subseteq B$

But $A' \subseteq p^{-1}(p(A'))$ is only true (for $A'=X$) if $p$ is a total function.

And then it gets hard to find any further properties absent for partial functions

  • $p(A'\cap B') \subseteq p(A')\cap p(B')$ and $p(A'\cap B')=p(A')\cap p(B')$ for injective $p$
  • $p(A'\cup B')=p(A')\cup p(B')$
  • $p(A' \setminus B') \supseteq p(A') \setminus p(B')$ and $p(A' \setminus B')=p(A') \setminus p(B')$ for injective $p$
  • $p(A' \operatorname{\Delta} B') \supseteq p(A') \operatorname{\Delta} p(B')$ and $p(A' \operatorname{\Delta} B')=p(A') \operatorname{\Delta} p(B')$ for injective $p$
  • $p(A'^c) \subseteq p(A')^c$ for injective $p$
  • $p(A'^c) \supseteq p(A')^c$ for surjective $p$

In a certain sense, the above considerations only identified two properties ($p^{-1}(A^c)=p^{-1}(A)^c$ and $A' \subseteq p^{-1}(p(A'))$) of total functions absent for partial functions, since $p^{-1}(Y)=X$ is just a restatement of the property for being total. Another absent property is suggested at the and of this answer, which is related to operators $\operatorname{op}:\mathcal{P}(Y)\to\mathcal{P}(Y)$ satisfying $A \subseteq B \Rightarrow \operatorname{op}(A) \subseteq \operatorname{op}(B)$. My question is which other properties of this type (i.e. properties which don't explicitly mention elements) are absent for partial functions.


3 Answers 3


This answer works out the absent property suggested at the the and of this answer.

Let $f: X \to Y$ be a total function. Let $\operatorname{op}_x:\mathcal{P}(X) \to \mathcal{P}(X)$ and $\operatorname{op}_y:\mathcal{P}(Y) \to \mathcal{P}(Y)$ be increasing operators, i.e. satisfying $A \subseteq B \Rightarrow \operatorname{op}_x(A) \subseteq \operatorname{op}_x(B)$ for all $A,B\subseteq X$ and the analogous condition for $\operatorname{op}_y$. Then the following two conditions are equivalent:

  1. $f(\operatorname{op}_x(A)) \subseteq \operatorname{op}_y(f(A))$ for all $A \subseteq X$
  2. $\operatorname{op}_x(f^{-1}(B)) \subseteq f^{-1}(\operatorname{op}_y(B))$ for all $B \subseteq Y$

The proof uses $A \subseteq f^{-1}(f(A))$ and $f(f^{-1}(B)) \subseteq B$ for both directions.
1. -> 2.: $\operatorname{op}_x(f^{-1}(B)) \subseteq f^{-1}(f(\operatorname{op}_x(f^{-1}(B)))) \subseteq f^{-1}(\operatorname{op}_y(f(f^{-1}(B)))) \subseteq f^{-1}(\operatorname{op}_y(B))$
2. -> 1.: $f(\operatorname{op}_x(A)) \subseteq f(\operatorname{op}_x(f^{-1}(f(A)))) \subseteq f(f^{-1}(\operatorname{op}_y(f(A)))) \subseteq \operatorname{op}_y(f(A))$

For a partial function $p:X\to Y$, we only have $A\cap p^{-1}(Y) \subseteq p^{-1}(p(A))$ and $p(p^{-1}(B)) \subseteq B$. Hence the same proof only yields something weaker for partial functions.
1. -> 2.: $\operatorname{op}_x(p^{-1}(B))\cap p^{-1}(Y) \subseteq p^{-1}(p(\operatorname{op}_x(p^{-1}(B)))) \subseteq p^{-1}(\operatorname{op}_y(p(p^{-1}(B)))) \subseteq p^{-1}(\operatorname{op}_y(B))$
2. -> 1.: $p(\operatorname{op}_x(A\cap p^{-1}(Y))) \subseteq p(\operatorname{op}_x(p^{-1}(p(A)))) \subseteq p(p^{-1}(\operatorname{op}_y(p(A)))) \subseteq \operatorname{op}_y(p(A))$
So we only have that the following two conditions are equivalent:

  1. $p(\operatorname{op}_x(A\cap p^{-1}(Y))) \subseteq \operatorname{op}_y(p(A))$ for all $A \subseteq X$
  2. $\operatorname{op}_x(p^{-1}(B))\cap p^{-1}(Y) \subseteq p^{-1}(\operatorname{op}_y(B))$ for all $B \subseteq Y$

Finally, we need two counter examples to show that the unmodified conditions don't imply each other in either direction. Let $X=Y=\{1,2\}$ and let $\operatorname{op}_x=\operatorname{op}_y$ map $\{2\}$ to $\{1,2\}$ and map any other subset to itself. If $p$ is undefined at $1$ and maps $2$ to $2$ then condition 1. is satisfied and condition 2. is violated. If $p$ maps $1$ to $1$ and is undefined at $2$, then condition 2. is satisfied and condition 1. is violated.

  • A partial function is an epimorphism if and only if it is surjective, no need to be total.

But a partial function is a monomorphism (in the category of sets and partial functions) if and only if it is an injective total function. Here, a partial function $i:X\to Y$ is a monomorphism if it has the following cancelation property with respect to composition $$\forall W \forall h_1,h_2:W\to X\ [(\forall w\in W:i(h_1(w))=i(h_2(w)))\ \Rightarrow\ h_1=h_2]$$


For a partial function $p:X\to Y$, we have $p^{-1}(A_{1}\cap\ldots\cap A_{r})=p^{-1}(A_{1})\cap\ldots\cap p^{-1}(A_{r})$ if $r\geq 1$. But for $r=0$, this is only true of $p$ is total. Hence $$\begin{array}{c}A_{1}\cap\ldots\cap A_{r}\ \subseteq\ B_{1}\cup\ldots\cup B_{s} \\ \Downarrow\\ p^{-1}(A_{1})\cap\ldots\cap p^{-1}(A_{r})\ \subseteq\ p^{-1}(B_{1})\cup\ldots\cup p^{-1}(B_{s})\end{array}$$ can fail if $p$ is not total and $r=0$.

I missed this initially, especially in the post which triggered the question.


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