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First of all apologies because I'm not a mathematician. So let's suppose I have a number of sets $A_1,...,A_n$ and functions $f_i\colon A_i \to A_{i+1}$ with $1\le i<n$. I want to know how to notate a mapping exists from $A_1 \to A_n$ through the composition of partial functions $f_1\circ...\circ f_{n-1}$.

Image of the above problem

Note that in the above image there is a mapping from $a_{11} \to a_{42}$, $a_{12} \to a_{42}$, $a_{14} \to a_{44}$ but not from $a_{13}$ to any element in set $A_4$.

The context in which I would like to check such a mapping is so I can use it as a condition perhaps like so:

$f(a) = (f_1 \circ...\circ f_{n-1})(a) \\ f(a) = f \colon A^\prime_1 \to A_n \\ A^\prime_1 \subseteq A_1$

Note $A^\prime_1$ is used because $f(a)$ is a partial function.

EDIT before I used $\exists (f_{n-1}\circ ...\circ f_1)(a): A_1\rightarrow A_n$ in the if part of the conditional so I changed it to another example I could think of:

$b = \begin{cases}1 &\text{if } a \in A^\prime_1 \\ 0 &\text{otherwise} \end{cases}$

Questions:

  1. Is there some standard notation for checking such a mapping?
  2. If not, has what I written made sense or are there some more familiar or appropriate methods I could use?
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    $\begingroup$ If $X, Y$ are sets, then there is a function from $X$ to $Y$ iff either $X$ is empty or $Y$ is nonempty; that is, the only way there isn't a function from $X$ to $Y$ is if $X$ is nonempty and $Y$ is empty. So it's easiest to define $a$ as "$a=1$ unless $A_1\not=\emptyset$ and $A_n=\emptyset$ in which case $a=0$." (Perhaps you wanted to include additional conditions that the map $f$ is supposed to satisfy?) $\endgroup$
    – Noah Schweber
    Commented Apr 2, 2018 at 12:50
  • $\begingroup$ Yes you're right, I will edit the question $\endgroup$
    – Jonathan
    Commented Apr 2, 2018 at 14:36
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    $\begingroup$ You wrote "$f_i : A_j \to A_{j+1}$". Did you perhaps mean "$f_j : A_j \to A_{j+1}$"? That's what it seems you meant in the context of the formula for $b$. $\endgroup$
    – Lee Mosher
    Commented Apr 2, 2018 at 15:10
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    $\begingroup$ Note that your functions are partial - e.g. in your example, you don't in fact have a function from $A_3$ to $A_4$ since it's not defined on all elements of $A_3$. I think what you mean to ask is: "Suppose $f_i$ is a partial function from $A_i$ to $A_{i+1}$ for $1\le i<n$, and $a\in A_1$. How do we express "$(f_n\circ f_{n-1}\circ ...\circ f_1)(a)$ is defined" concisely?" Is this right? $\endgroup$
    – Noah Schweber
    Commented Apr 2, 2018 at 17:03
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    $\begingroup$ What you've written - "$\exists (f_{n-1}\circ ...\circ f_1)(a): A_1\rightarrow A_n$" - is not well-formed. The partial function $f_{n-1}\circ ...\circ f_1$ exists and is a partial function from $A_1$ to $A_n$, by definition. What you want to write is something like "$\exists x((f_{n-1}\circ ...\circ f_1)(a)=x)$" or similar. $\endgroup$
    – Noah Schweber
    Commented Apr 2, 2018 at 19:09

1 Answer 1

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The symbol "$\downarrow$" is often used to indicate defined-ness of a partial function on a given input. So e.g. you could abbreviate the condition by "$f_{n-1}\circ...\circ f_1(a)\downarrow$." I don't know of any shorter expression for it.

(Incidentally, if this is a condition you're interested in it may be worth doing a bit more work and letting (say) $g_{i,j}$ for $i\le j$ denote the function $f_{j-1}\circ...\circ f_i$; then your condition is just $g_{1, n}(a)\downarrow$. Introducing these new functions may or may not be worthwhile depending on what you're interested in.)

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