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My guess is that both are $\emptyset$ because if $g\circ f=\{(x,z)\mid \exists y\in \text{Im}f:(x,y)\in f\land (y,z)\in g\}$ then if $f$ or $g$ are the empty set then it doesn't exist any $y$ with such condition, so the set is empty. Am I right?

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    $\begingroup$ +1 I've never seen such composition with the empty set..$\emptyset\circ f$, where did you see it? $\endgroup$
    – user486983
    Oct 19, 2018 at 18:45
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    $\begingroup$ Yes, you are right. $\endgroup$ Oct 19, 2018 at 18:50
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    $\begingroup$ @Isah I need it for a project that I'm working on. $\endgroup$
    – Garmekain
    Oct 19, 2018 at 18:56
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    $\begingroup$ I'm a little confused but I think you are right? You are talking about function compositions so you have to talk about 2 functions.$\emptyset$ usually stands for the empty set, which isn't a function.Briefly looking this up, I found something about "empty functions" and empty set stuff but I didn't really read the material. (as an aside your definition of the composition $g\circ f$ is a little strange to me, but maybe I'm not familiar with the notation.It sounds like you are talking about the graph of $g \circ f$. I've never seen $(x,y) \in f$ before because $f$ isn't a set - but I understand) $\endgroup$
    – DWade64
    Oct 19, 2018 at 21:03
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    $\begingroup$ @DWade64 The empty set is indeed a function. $\endgroup$
    – Garmekain
    Oct 19, 2018 at 21:34

2 Answers 2

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Yes, you are right.

Under the convention that a function is a set $f$ of ordered pairs such that

if $(x,y)\in f$ and $(x,z)\in f$, then $y=z$

we can define function composition in the following way.

Let $f$ and $g$ be functions; then $g\circ f=\{(x,z):(x,y)\in f\text{ and }(y,z)\in g,\text{ for some }y\}$ is a function.

The empty set is obviously a function in the sense described above and, for every function $f$, $$ \emptyset\circ f=\emptyset=f\circ\emptyset $$ because $(a,b)\in\emptyset$ is false for every $a$ and $b$.

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This is old, but I don't think I agree with the accepted answer.

In the usual interpretation of $g\circ f$, we assume that $\mathrm{cod}(f) = \mathrm{dom}(g)$, or at least $\mathrm{Im}(f) \subseteq \mathrm{dom}(g)$. Then $g\circ f:\mathrm{dom}(f)\to\mathrm{cod}(g)$. If we don't have $\mathrm{Im}(f) \subseteq \mathrm{dom}(g)$, then we must regard $g$ as a partial function to take the composition.

Since the empty function has domain $\emptyset$, I claim we have $$ f\circ\emptyset = \emptyset, $$ but $$ \emptyset\circ f\ \textrm{is undefined (unless $f=\emptyset$)}. $$ Taking instead the laxer definition of composition where we disregard domains and codomains, I of course agree with egreg, and the claim would be equivalent to $$ \emptyset\circ_{\mathrm{lax}} f := \emptyset\circ \left(f\upharpoonright_{f^{-1}(\mathrm{dom}\ \emptyset)} \right) = \emptyset\circ\emptyset = \emptyset. $$

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