So I think that if there exists two functions $f_1 \colon X \to Y$ and $f_2 \colon Y \to Z$ you can notate their composition using $(f_1 \circ f_2)(x) = f_3(x)$ right? If I have many functions that I need to compose, is there some shorter notation I can use such as $f_3 \colon X \to Y \to Z$ or something? The first example can get tedious when there are a lot of functions/mappings.

I already thought of putting all the sets in a some auxiliary set $V = (X_1,...X_n)$ and then defining $f_i \colon X_i \to X_{i+1}$ with $0 < i \leq n$ so I can notate composition like so $(f_1 \circ...\circ f_n)(x)$. But some functions are partial so I need to explicitly use the partial function arrow with a bar sometimes in this format $f_3 \colon X \to Y \to Z$. Would have demonstrated the bar arrow here but needs extra package installed.

Edit: In the context of the paper I'm writing, I'm explaining the specific ways in which the functions work with words because its computer science work that is hard to express succinctly with maths notation, but I would still like to show the relationship between the these functions using function notation. I assumed this was the best way to show this relationship. I'm open to the suggestion of alternative notations though.

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    $\begingroup$ I'm not sure what you think could be shorter than $f_n\circ\cdots\circ f_1$. Somehere you have to specify which functions it is you're composing! $\endgroup$ Apr 3, 2018 at 0:52
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    $\begingroup$ You need to refer to $f_1, \ldots, f_n$ at some point in your notation. Specifically, since the notation "$X \to Y$" cannot specify the particular function $f_1$, why do you think "$X \to Y \to Z$" can refer specifically to $f_2 \circ f_1$? (Note also that you have the order of composition backwards.) $\endgroup$
    – angryavian
    Apr 3, 2018 at 0:54
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    $\begingroup$ $\mathop{\bigcirc}\limits_{i=1}^n f_i$, perhaps? I'm not sure that's much more readable... $\endgroup$ Apr 3, 2018 at 0:56
  • $\begingroup$ @angryavian In the context of what I'm writing I'm explaining the functions with words because its more computer science work that is hard to express in a non-confusing way with maths notation, but I would still like to show the relationship between the sets using the notation. So I thought in the context it might be obvious that each $\to$ would imply the function I had previously explained that maps the two sets. I'll make an edit and specify this. $\endgroup$
    – Jonathan
    Apr 3, 2018 at 1:09
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    $\begingroup$ In some contexts you could get away with $f_n f_{n-1}\ldots f_2 f_1$, but this had better be a context where you're not using juxtaposition for anything else (or else some other convention makes it clear when juxtaposition means composition and when it means whatever else). $\endgroup$ Apr 3, 2018 at 1:12

2 Answers 2


You don't need to have to add the $(x)$ in the example you give. So you can write $(f_1 \circ f_2)(x) = f_3(x)$ as $f_1 \circ f_2 = f_3$.

If you are doing a large indexed set of composed functions then big circle notation works just like $\Sigma$ for sums and $\Pi$ for product. $\bigcirc_{i=1}^nf_i$

However if you are doing a great deal of compositions simply define your notation to have juxtaposition to be equivalent to composition $f_2(f_1(x)) = (f_1f_2)(x) $ (or whatever ordering convention is most useful to you).


The standard notation in my area of interest (fractal geometry and analysis on fractals) is to use superscript notation for functional composition. This is not unreasonable, as the set of functions on a space, together with the operation of composition, has a nice algebraic structure (a monoid? a semigroup? I can never remember which is which—there is an identity function, but inverses don't necessarily exist).

To compose multiple functions, define a "word" as a sequence (or tuple) of indices, and denote the composition along that sequence with a superscript. For example, $$ \omega = (\omega_1, \omega_2, \omega_3, \omega_4, \omega_5), $$ where each of the $\omega_i$ is an element of some index set. Then $$ f^\omega = f_{\omega_5} \circ f_{\omega_4} \circ f_{\omega_3} \circ f_{\omega_2} \circ f_{\omega_1}. $$


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