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I understand that for composite function $(g\circ f)(x)$ to exist, range of $f(x)$ must be a subset of domain of $g(x)$. This is so that every output value of $f(x)$ is mapped to one value of $g(x)$.

However, is this on the assumption that both $g(x)$ and $f(x)$ are functions? Must they both be functions?

Imagine a case where $f(x) = \pm \sqrt{x}$ and $g(x) = x^2$. in this case $g(f(x))$ is a function right? So I suppose we do not need both $g(x)$ and $f(x)$ to be functions?

So what does "range of $f(x)$ must be a subset of domain of $g(x)$" actually conclude?

Addon: I've realised that the example I have given, is for the special case when $g$ and $f$ are inverse of one another. so $(g\circ f)(x) = x$. My conclusion is that for composite function $(g\circ f)(x)$ to exists, either

  1. $g(x)$ and $f(x)$ has to be functions, or
  2. $g(x)$ and $f(x)$ are inverse of one another.

Am I right? Any help is much appreciated! I still need help.

Yan Bo Pei
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  • Considering the concept of relation (a.k.a. multivalued function), an obvious and essentially the only criterion for $g \circ f$ to become a function is that $$ \text{for each $x$, the value of $g(y)$ is the same for any possible value $y$ of $f(x)$}. $$ If $g$ is a function and $f$ is the inverse of $g$ (so that in general $f$ is not a function but a multivalued function), then this is obviously satisfied. But there is a plethora of other examples which also make $g \circ f$ a function while $f$ and $g$ have no obvious connection. – Sangchul Lee Sep 22 '18 at 16:19
  • in my case, im looking if firstly g and f are functions.
    1. they are both functions.

    1.1) For g(f(x)) to be functions, range of f must be subset of domain of g.

    1. they are not both functions.

    2.1) Can g(f(x)) still be a function? 2.1.1) Yes if g and f are inverse of one another. g(f(x)) = x and therefore a function. 2.1.2) If g and f are not inverse of one another. can they still be functions? Are there any conditions (as a rule of thumb) for them to be functions?

     – Yan Bo Pei Sep 22 '18 at 17:30

1 Answers1

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I'd hate to 'post' this as an 'answer' because it's not really an answer per say but something I would've wrote as a comment over an 'answer.' (Because I'm just thinking out loud)

Usually when I see statements of compositions of functions g$\circ$f, they usually start as if g: X $\rightarrow$ Y is surjective and f:Y $\rightarrow$ Z is surjective, then g $\circ$ f: X $\rightarrow$ Z is surjective. It's in an "if-then" form. So consider a statement: If g: X$\rightarrow$ Y is a function and f: Y $\rightarrow$ Z is a function, then the composition g $\circ$f: X $\rightarrow$ Z is a function. Well with logic for the P implies Q table, if P is false, then the statement is true. Hence, if g or f are not functions, then by that, g $\circ$ f would be a function would be a valid/true statement. The contrapositive would also say that if g$\circ$f is not function, then g is not a function or f is not a function-- again just rambling off ideas.

Also, I'd be careful when saying 'inverses.' If a function f:X $\rightarrow$Y has an inverse say f$^{-1}$:Y $\rightarrow$ X, then that implies f$^{-1}$ is a function, but as you mentioned above, the squareroot is not a function. If anything, might be able to consider it as the pullback.

Mind my loose words, I too am trying to be more concise, but I hope my comment gets you thinking.

Ren
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  • Interesting thoughts ... however, based on your logic of P implies Q, if P is false, the statement Q is only vacuously true. – Yan Bo Pei Sep 22 '18 at 17:37
  • just considering the two cases if P is false, then the statement is true, so Q could be false or true, so g$\circ$f may be a function or may not, but yeah I think the example from D_S is just showing that a composition of a well defined function and one that is not does give a well defined composition function-- implying that maybe for a short answer it'd be no, if g$\circ$f is a function, what can you imply about f and g-- at least in the specific example, one was a function, one wasn't. – Ren Sep 22 '18 at 18:54