Allen Knutson's Math 113, spring '04, HW #1, due 1/29

1. How many isomorphisms are there between these two graphs?

There are 8. It's easy to prove that a vertex of "degree 3" (meaning, there are 3 edges coming out) must be corresponded to another vertex of degree 3. So we first choose the 2! ways to correspond the left guy's two degree-3 vertices with the right guy's. Then we have to correspond the hangers-on to the hangers-on, and at each end we have 2! such choices. In all, 2*2*2 = 8.

2. Find an isomorphism from the first graph to the second graph:

There are so many, that if you start anywhere, you are unlikely to get stuck.

3. Let G,H be two graphs with 5 vertices and 9 edges. Show they are isomorphic. How many isomorphisms are there from G to H?
If all vertices in G were connected to one another, that would be 10 edges. So exactly one edge is missing. Let's split up the vertices of G into the two that don't like each other, a,b, plus the remaining three friends c,d,e who are sad that their two friends are having a tiff. Meanwhile in H there is also a pair of antipathetic vertices p,q and three gregarious ones r,s,t.

Now any correspondence between a,b and p,q (of which there are 2!) and any correspondence between c,d,e and r,s,t (of which there are 3!) will take the unique non-edge to the unique non-edge, so there are 2!*3!=12 such isomorphisms. In particular, 12>0 so there _are_ isomorphisms.

4. Prove that the number of 1:1 functions from a k element set to a (k+n)-element set is (n+1)(n+2)...(n+k). (Hint: What should you induct on?)
If k=1, we're injecting a 1-element set into an n+1 element set, and there are indeed n+1 ways.

More generally, call the number we're trying to compute F(n,k). If k>1, let's first choose where one element goes -- that's n+k possibilities -- and then for each choice we have to take the remaining k-1 different elements to the remaining (n+k)-1 = n+(k-1) different elements (not n+k, because we want the function to be 1:1. So we get an equation

F(n,k) = (n+k) F(n,k-1).

BY INDUCTION, we know what F(n,k-1) is; it's (n+1)....(n+(k-1)). That together with the above equation gets us the formula we want for F(n,k).

If you said "and so on" in your proof, you were almost certainly hiding an induction on k there. Which is fine, but you should realize it.

5. Use some symmetries to show that the number of partitions of {a,b,c,d,e} is 52. Please, please do something more systematic than writing out all 52.
There's 1 way to break into one group {{a,b,c,d,e}}.

If we break into 2 groups, we could get 5=1+4 in 5 ways, e.g. {{a},{b,c,d,e}}, and 5=2+3 in 5*4/2 ways, e.g. {{a,b},{c,d,e}}.

If we break into 3 groups, we could get 5=1+1+3 in 5*4/2 ways, e.g. {{a},{b},{c,d,e}}. Or 5=1+2+2 in 5*3 ways, e.g. {{a},{b,c},{d,e}}.

If we break into 4 groups, we get 5=1+1+1+2 in 5*4/2 ways.

In 5 groups, the only way is with each one in her own group.

So 1+5+10+10+15+10+1=52.

6. Prove that a function is an epimorphism iff it's onto. Be as rigorous as you can, and make sure to prove both directions!
Onto => epi. f:X->Y is onto, want to show that for all g1,g2:Y->Z, g1 o f = g2 o f implies g1=g2.

For all y\in Y, pick an x such that y = f(x), using ontoness. Then g1 o f (x) = g2 o f(x) = g1(y) = g2(y). Since this was for all y\in Y, g1=g2. QED.

Not onto => not epi, want to show that there exists a bad g1,g2 with g1 o f = g2 o f but not g1=g2.

Let b \in Y such that no x\in X has f(x)=b, using not-ontoness. Let Z = {1,2}. Define g1:Y->Z by g1(y)=1 for all y\in Y. Define g2:Y->Z by g2(b)=2, but g2(y)=1 for all y \in Y other than b.

Then for all x\in X, g1(f(x)) = 1 = g2(f(x)), since f(x) is not b. So g1 o f = g2 o f. But g1(b) = 1, g2(b) = 2, so g1 is not g2. QED.