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What type of characteristics should properties have for the following to hold true?

If $f,g$ each have property $p$ then $f \circ g$ also has property $p$

Examples:

If $f,g$ continuous then $f \circ g$ is continuous

If $f,g$ entire then $f \circ g$ is entire

If $f,g$ differentiable then $f \circ g$ is differentiable

If $f,g$ one to one then $f \circ g$ is one to one

If $f,g$ Contractible then $f \circ g$ is Contractible

If $f,g$ polynomial then $f \circ g$ is polynomial

If $f,g$ lineaer then $f \circ g$ is linear

Counter Examples:

If $f,g$ integrable then $f \circ g$ is integrable (false) see Robert Israel's answer

If $f,g$ measureable then $f \circ g$ is measureable

Update Edit: The motivation for the question was to use the statement as a filter to specify properties p, I had seen too often question being asked for a specific property, was wondering if there is a way to devise a test for p rather than test each p individually to see if it satisfies the statement. (This is as good as I can explain my intent for asking the question, making a list seems to be a beneficial side effect). If anyone can elucidate the motivation with better mathematical terminology please edit.

jimjim
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  • What assumptions do you make about their domains/codomains? – platty Aug 04 '17 at 22:03
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    Could you give an explicit example for your counterexample? – mfl Aug 04 '17 at 22:03
  • @mfl : woops, that is wrong, fixing the question – jimjim Aug 04 '17 at 22:06
  • @platty : assumptions that make the statement true, trying not to restrict it yet, but if some assumptions help, then I'll take them. – jimjim Aug 04 '17 at 22:09
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    Someone who knows more category theory than me can probably give a decent answer, but generically in some category $\mathcal{C}$ the morphisms will generally preserve some structure. It follows that the composition of morphisms preserves that structure. This will go in your "examples" group, and includes things like linearity, measurability, (group/ring) homomorphism – Mark Schultz-Wu Aug 04 '17 at 22:10
  • @Mark : thank you, now I know what I am looking for. – jimjim Aug 04 '17 at 22:11
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    Yes, these properties all define sub-categories with the same objects but a subset of maps. (They also have the property that the identity has the property.) An example that wouldn't have the identity is one-to-one but not-onto functions, which would thus not indicate a sub-category. – Thomas Andrews Aug 04 '17 at 22:12
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    Simple examples: constant and polynomial. Also Lipschitz, contraction, and uniformly continuous – Hagen von Eitzen Aug 04 '17 at 22:14
  • More basic pro example without identity: Functions which are not one-to-one. (Or, conversely, functions which are not onto.) These two examples are more like ideals, in that if $f$ is not one-to-one, then $g\circ f$ is not one-to-one for any $g$, and dually if $f$ is not onto, then $f\circ g$ is not onto for any $g$.) – Thomas Andrews Aug 04 '17 at 22:18
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    also simple examples, linear, invertible (when you think of the category of linear maps) I think you really need to add more constraints to stop the problem from being too open. – mdave16 Aug 04 '17 at 22:20
  • Pro example, again without (all) identity functions: Contractible. This is a left-right ideal example - if $f$ is contractible to a point, then $f\circ g$ and $g\circ f$ is contractible to a point for any $g$. – Thomas Andrews Aug 04 '17 at 22:27
  • @mdave16 : now I realise that too, maybe instead of constraints I can separate them, but then that would make it about the specifity or property not what properties will pass the filter of statement being true. – jimjim Aug 04 '17 at 22:28
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    Any function that factors through some set in a class. For example, if the class is "finite," then $f:A\to B$ factors through the class if $f:A\to C\to B$ for some pair of functions $A\to C,C\to B$ with $C$ finite. This is, again, a left-right ideal example, since if $f$ factors through the class, then $f\circ g$ and $g\circ f$ factor through that class, for any $g$. (The constant example is just the case where the class contains singleton sets.) – Thomas Andrews Aug 04 '17 at 22:36
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    @ThomasAndrews, would you mind putting all your comments into one organised answer? That way, I can upvote and it makes a difference. :) – mdave16 Aug 04 '17 at 23:31
  • @ThomasAndrews : mdave16 said, I too can upvote it and set it as accepted answer. – jimjim Aug 04 '17 at 23:38

2 Answers2

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"If $f, g$ integrable then $f \circ g$ is integrable." - False!

For example, $f(x) = g(x) = e^{-x}$ is integrable on $(0,\infty)$, but $f(g(x)) = e^{-e^{-x}}$ is not integrable there.

Robert Israel
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  • Thank you. I updated my question using this info, apologies it now makes this answer not look relevant – jimjim Aug 04 '17 at 22:25
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When all identity maps satisfy your property, then this is, at heart, a sub-category, where the objects are the same and we take a subset of the morphisms.

You probably want to confine yourself to cases where all identity maps satisfies your property, because the cases where some identity functions are not in the set are "odder."

For example:

  1. Functions which are not onto.
  2. Functions which are not 1-1.
  3. Functions which are contractible to a point.
  4. Functions that factor through a class of sets - $f:A\to B$ factoring though the class if there is a $C$ in the class such that $f$ factors as $A\to C\to B$ for some functions.
  5. The case of constant functions is a special case of 4.

These cases also have a side-property that give $f,g$, for $g\circ f$ to have the property, you need only one of them to have the property. For (1), you only need $g$ in the class. For (2), if $f$ is in the class. For (3-5), If either $f$ or $g$ satisfies our property, then $g\circ f$ is.

So, these cases are sort of like "ideals" in the parent category, rather than sub-categories.

The cases (1) and (2) act like prime ideals - if $g\circ f$ is not one-to-one, then one of $f$ or $g$ is not one-to-one.

If you have a collection of these properties (either with the identities or not,) then the intersection of them is also a property like this. So differentiable one-to-one functions are closed under composition.

Thomas Andrews
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