mps's New Writeups

discontinuous identity map (idea)

2001-10-24T03:15:23Z

Dropping excess symbols when writing mathematics is a common practice (when was the last time you read a proof that began: let (R,+,*) be a ring?). But when we try to let notation do the thinking for us this practice may lead us into error. For example:

A topological space, by definition, consists of a set X and a topology T on X, and so technically should be denoted (X,T). Writing that all the time, however, wastes precious paper, so such a space is usually denoted X.

The trouble comes in when we talk about continuous maps of topological spaces. For a map f between topological spaces to be continuous, it must be that the preimage of a set open in the codomain's topological space is open in the domain's…

identity map (idea)

2001-10-24T03:12:17Z

Definition: Let X be a set. Then the identity map on X is the map id:X->X defined by id(x)=x.

The identity map is a special case of the inclusion map.

preimage (idea)

2001-10-24T03:06:31Z

Definition: Let f :X->Y be a map and A a subset of Y. Then the preimage of A under f, denoted f ^-1(A), is equal to {x in X | f (x) in A }.

That is, the preimage of a set A under a map f is the set of elements of the domain of f that get mapped to an element of A. For example, the preimage of {1} under f :R->R defined by f (x)=x² is {-1,1}, because (-1)²=1 and 1²=1, and for no other x is this true.

Psalm 115 (idea)

2001-10-23T00:36:43Z

(composite translation)

Not to us, YHVH,
but to your name be the glory,
because of your love and faithfulness.

Why do the nations say,
"Where is their Elohim?"

Our Elohim is in the heavens;
he has done whatever he has pleased.

Their idols are silver and gold,
made by human hands.

They have mouths, but do not speak;
eyes, but do not see;

they have ears, but do not hear;
noses, but do not smell;

they have hands, but do not feel;
feet, but do…

short five lemma (idea)

2001-10-21T22:48:00Z

Theorem. (The short five lemma) Let 1 -> A -> B -> C -> 1 and 1 -> A' -> B' -> C' -> 1 be short exact sequences, and suppose that the diagram

        a     b
1 -> A  -> B  -> C  -> 1
    f|    g|    h|
     v     v     v
1 -> A' -> B' -> C' -> 1
        a'    b'

commutes. Then:

If f and h are monomorphisms, then g is a monomorphism.
If f and h are epimorphisms, then g is an epimorphism.
If f and h are isomorphisms, then g is an isomorphism.

Proof.

Suppose g(x)=1. The diagram commutes, so h(b(x))=b'(g(x))=1. h is injective; therefore, b(x)=1. Since x is in ker b, there is y in A such that a(y)=x. Thus a'(f(y))=g(a(y))=1. But a' is injective (since ker a' =

…

commutative diagram (idea)

2001-10-21T20:24:56Z

The basic commutative diagrams are:

A -> B       A -> B
|    |  and    \  |
v    v          v v
C -> D            C

To say that a diagram is commutative means the following. Suppose x is in A, and let A -> C -> ... -> B and A -> D -> ... -> B be two ways of chasing arrows, or paths, around the diagram (only going forward on the arrows). Then if the first path leads to y in C, the second path leads to the same y.

For example, let f be a map from A to B, g be a map from B to C, and h be a map from A to C. If the diagram

A -> B
  \  |
   v v
     C

commutes, then for any x in A, h(x)=g(f(x)).