0
$\begingroup$

I want to write a proof to show that the composition of two functions is a function. To do that, I need to define the composition of two functions, but I haven't been able to verify my attempt. Here it is.

Let $f : X \rightarrow Y$ and $g : Y \rightarrow Z$ be well-defined functions, and let $f \circ g : X \rightarrow Z$ be the composition of $g$ to $f$. The composition $f \circ g$ is defined as

$$f \circ g := \{(x, z) : \exists y \ni (x, y) \in f \, \mathrm{and} \, (y, z) \in g\}$$

Please I'd like to know it this definition is correct, and if not, the correct definition. Thanks.

$\endgroup$
2
  • 1
    $\begingroup$ See Composition of relations. $\endgroup$ Commented Sep 13, 2023 at 8:01
  • $\begingroup$ Thanks for the link @MauroALLEGRANZA. $\endgroup$ Commented Sep 13, 2023 at 8:18

1 Answer 1

Reset to default
0
$\begingroup$

I assume you understand the meaning of composite functions, and that you're interested in the explicit definition.
In that case, to define it formally it is best to understand that when composing the functions, although $g$ acts on $Y$, the composition of $g$ on $f$ acts on $Image(f) = Im(f) := \{f(x) | x\in X\} \subseteq Y$.

So the composite consists of all the copules $(x,z)$ such that x is sent by f to some image $y=f(x)$ and $g$'s action on that image results in $z = g(y)$.

$ g∘f:=\{(x,z) |\ \exists y \in Y : f(x) = y \land g(y) = z \}$

$\endgroup$
7
  • $\begingroup$ Oh I see. So my definition above was for $g \circ f$? $\endgroup$ Commented Sep 13, 2023 at 8:19
  • $\begingroup$ Yes :) That is why I emphasised the intuition from which the formal definition is derived. $\endgroup$ Commented Sep 13, 2023 at 8:21
  • $\begingroup$ So if I were to write the reverse, I wouldn't be able to use the same domain and range as I used in the post? $\endgroup$ Commented Sep 13, 2023 at 8:23
  • $\begingroup$ In your post you stated that $f:X \rightarrow Y$ and $ g: Y \rightarrow Z$, so the only logical composition is of g on f, since f cannot be composed on g (if X is not equal to Z). $\endgroup$ Commented Sep 13, 2023 at 8:25
  • $\begingroup$ I see. So would this be a better representation? Let $f:X_{1} \rightarrow Y_{1}$ and $g:X_{2} \rightarrow Y_{2}$ be arbitrary and well-defined. Then $f \circ g := \{(x, z) : \exists y \ni (x, y) \in g \, \mathrm{and} \, (y, z) \in f\}$ $\endgroup$ Commented Sep 13, 2023 at 8:31

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.