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Why are inverse images of functions more central to mathematics than the image?

I have a sequence of related questions:

  1. Why the fixation on continuous maps as opposed to open maps? (Is there an epsilon-delta definition of open maps in metric spaces?)

  2. Is there an inverse-image characterization of homomorphisms in algebraic categories? (What kind of map do you get if you look at a map from a group to another group, where inverse images of subgroups are subgroups?)

  3. Inverse images have better set-theoretic properties than the image (for instance, commuting with unions, intersections, etc..) This clearly is a direct consequence of definition of a function. There is an asymmetry in the definition of a function (the domain and codomain behave differently with respect to the function). I think this also has consequences for differences between existence and uniqueness of left and right inverses for one-to-one or onto functions. Why this asymmetry? What are the historical reasons for the asymmetry? Whats sort of mathematics do we have if the definition of a function was purely symmetric? (For instance, f(a) may give multiple values, just like f^-1(a) may have multiple values).

  4. Is it accurate to say the definitions for monomorphisms and epimorphisms in category theory correct for the asymmetry? (And hence, the notion of epimorphisms and onto-morphisms in concrete categories don't coincide)

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This is a good community wiki candidate. –  Bill Kronholm Apr 27 '10 at 0:45
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The symmetrically defined functions you mention in 3. are just relations. –  Qfwfq Apr 27 '10 at 11:28
    
Would be great if someone could clean up the tags and retag this [soft-question]! –  Sonia Balagopalan Apr 27 '10 at 13:16
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4 Answers

up vote 6 down vote accepted

Open sets can be identified with maps from a space to the Sierpinski space, and maps out of a space pull back under morphisms. (In other words, if you believe that the essence of what it means to be a topological space has to do with functions out of the space, you are privileging inverse images over images. A related question was discussed here.) I think essentially this kind of reasoning underlies the basic appearances of inverse images in mathematics. For example, in the category of sets, subsets can be identified with maps from a set to the two-point set, and again these maps pull back under morphisms. This should be responsible for the nice properties of inverse image with respect to Boolean operations.

Your third question was asked, closed, and deleted once; I started a blog discussion about it here.

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Continuity is important not because of its inverse-image-ness, but because the definition corresponds to the geometric notion that it's intending to capture. A continuous function "takes close things to close things". This geometric intuition precedes the notion of open sets, and even the epsilon-delta definition of continuity.

I think the same sort of thing is true of functions in general. The asymmetry is there as a consequence of the idea that the definition is intending to capture; functions weren't originally thought of as a special class of relations, they were thought of as "rules" for manipulating numbers, and the idea of f(x) being unique and f${}^{-1}$(x) not being unique is the only natural way to capture the idea of a "rule" in a more general context.

I don't know that the notions of monomorphisms and epimorphisms really "correct" for the asymmetry, but I don't think it's something that ought to be corrected anyway. Monomorphisms of sets are the same as injective functions, and epimorphisms of sets are the same as surjective functions, and that's the way it should be. In categories like Ring where, say, epimorphisms aren't always surjective, it's because the non-surjective epimorphisms, in some sense, are "surjective as far as ring maps are concerned"; for example, the inclusion of $\mathbb{Z}$ into $\mathbb{Q}$ is epi because a map out of $\mathbb{Q}$ is determined by what it does to $\mathbb{Z}$. I don't think this has much to do with the asymmetry you're describing.

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I think that the first 2 paragraphs of this answer are truly excellent. –  teil Apr 27 '10 at 2:22
    
I understand that continuity is not important because of its inverse-image-ness, but because it corresponds to the geometric notion. But WHY does the inverse-image-ness correspond to this geometric notion? It seems that there are seemingly-unrelated different geometric notions that ALSO corresponds to inverse-image-ness (such as measurable functions). I suppose the question is, why the prevalence of inverse-image-ness? –  atonaltensor May 8 '10 at 17:22
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Questions 1, 3, and 4 have been very well explained in the other answers, but I have something to remark about Question 2.

Very frequently, objects that are meant to be like spaces will have some kind of algebraic data attached to them. But this algebraic data is attached contravariantly, that is, there's some functorial relationship between your category of objects and the opposite of the category of algebraic structures.

For example:

  • Sets and Boolean Algebras. The power-set functor mentioned in Sammy Black's answer actually gives a contravariant functor from sets to Boolean algebras. This functor actually embeds the category of sets into the opposite category of Boolean algebras, so sets may be regarded as Boolean algebras with certain properties, except the maps go the wrong way.

  • Schemes and Rings. A scheme is locally isomorphic to an object in the opposite category of commutative rings. In fact, the category of schemes admits a fully-faithful embedding into $Set^{Rng}$, the free cocompletion of $Rng^{op}$. This is called the "functor of points" approach to schemes.

  • Compact Hausdorff Spaces and Unital C*-Algebras. There's a contravariant equivalence between the category of compact Hausdorff spaces and the category of C*-algebras with unit.

  • Locales and Frames. A frame is a kind of distributive lattice, and is described in a completely algebraic way. It's space-like counterpart, called a locale, is studied in so-called "Pointless Topology" (don't laugh), and the category of locales is defined to be the opposite category of frames. This was inspired by the last example, which is:

  • Topological Spaces and their Lattices of Open Sets. To every topological space, there is associated a certain lattice (the lattice of open sets). The requirement is that this association be contravariantly functorial - that is, every map of topological spaces must give rise to a map of lattices in the opposite direction. And that's what we have: a continuous map is one that induces a well-defined inverse-image map taking open sets to open sets.

So the idea that open maps seem to be more straightforward (so to speak) than continuous maps may be a common one, but in fact it seems that we get better categories of spaces if we ask the algebraic data to be contravariant.

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1 Many important properties of topological spaces are preserved by continuous maps (but not necessarily open maps): connectedness and compactness come to mind immediately. But more importantly, the most familiar, natural maps that we can define are continuous, but not necessarily open: polynomials $\mathbb{R}^n \to \mathbb{R}^m$.

2 Inverse image of a subgroup under a homomorphism is a subgroup.

3 There is a contravariant functor from the category Set to itself, mapping a set $X$ to its power set $\mathcal{P}(X)$ and sending the morphism $f:X \to Y$ to the inverse image $f^{-1}:\mathcal{P}(Y) \to \mathcal{P}(X)$. A "purely symmetric" function would be a symmetric relation on $X \times Y$, no? Functions are, after all, relations with an extra property that deliberately breaks the symmetry!

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As for the 2, I think he's actually asking whether a map of sets, such that the inverse image of every subgroup is a subgroup, is actually a group homomorphism. –  Qfwfq Apr 27 '10 at 11:41
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If that's the question, the answer is no. Let $G$ be a cyclic group of prime order - then the requirement that a function $G\rightarrow G$ pull back subgroups to subgroups is very weak (it implies only that no nonzero element maps to zero), and most such functions won't be group homomorphisms. –  Owen Biesel Apr 27 '10 at 17:49
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