Given two functions $f:A\rightarrow C,\ g:B\rightarrow C$, is there a standard way to denote the function $h: A \sqcup B \rightarrow C$ such that $h\mid_A\equiv f$ and $h\mid_B\equiv g$? Is it something like $f\amalg g$ or $f\sqcup g$?

  • $\begingroup$ Yes, I think that these are quite reasonable notations. However, I doubt that these are standard. $\endgroup$
    – Crostul
    Nov 13, 2015 at 15:25
  • $\begingroup$ If you wrote $f \sqcup g$ everybody would probably know what you meant. $\endgroup$
    – user98602
    Nov 13, 2015 at 15:33
  • $\begingroup$ I'd use the same notation. However, there's also the combination of two functions $f:A\rightarrow C$ and $g:B\rightarrow D$ into a function $h:A\sqcup B\rightarrow B\sqcup D$, which might also be denoted by $f\sqcup g$. Does anyone have a good suggestion for a notation for this? $\endgroup$
    – Andi Bauer
    Apr 29, 2021 at 14:30

2 Answers 2


You can use the same notation as for sets. Not only because "everybody will know what you mean", but also because it is coincidentally well defined. This is because functions are sets of tuples (the set theoretic definition of functions). If you regard the functions $f$ and $g$ as sets, then the set $f\sqcup g$ is again a function and it is exactly the function you wanted to get. Indeed:

Writing $f$ and $g$ as sets yields $$f = \{(a,y)\mid a\in A \mbox{ and } y = f(a)\}$$ $$g = \{(b,y)\mid b\in B \mbox{ and } y = f(b)\}$$

Taking the ordinary disjoint union of sets yields $$f\sqcup g = \{(c,y)\mid (c\in A \mbox{ and } y = f(c)) \mbox{ or } (c\in B \mbox{ and } y = f(c))\}$$ which agrees with your definition of disjoint union of functions.


Literally the first thing I thought of when I saw the title of your post and the first bunch of words (prior to "standard") was: $f\sqcup g.$ Since I can't recall ever having seen it, I must agree with Crostful that it may not be "standard," per se. However, given the automaticity of that notation popping into my mind, I must also agree with Mike Miller that it is readily "standardizable."


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .