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Algebra (647), Homework 2, solutions.
1. I.7: 3.
We need to show that given a group H and maps fi : H → Gi there is a unique map f : H → G 1 × G 2 so that πi ◦ f = fi. We de ne f (h) = (f 1 (h), f 2 (h)). This makes it obvious that πi ◦ f = fi, so it gives existence. If g is a map to G 1 × G 2 so πig = fi then g(h) = (f 1 (h), f 2 (h)). So g = f. This gives uniqueness.
- I.7, 4. Here we need: given maps f 1 : A 1 → B, f 2 : A 2 → B there exists a unique map f : A 1 × A 2 → B such that f ◦ ιi = fi. We de ne f (a 1 , a 2 ) = f 1 (a 1 ) + f 2 (a 2 ). It's straightforward to check that f is a homomorphism. As far as the commuting diagram, f (ι 1 (a)) = f (a, 0) = f 1 (a) + f 2 (0) = f 1 (a) and similarly for f 2. This gives us existence. For uniqueness we note that for any g with g ◦ ιi = fi we have g(a 1 , 0) = g(ι 1 (a 1 )) = f 1 (a 1 ) and similarly for a 2. Also, (a 1 , a 2 ) = (a 1 , 0) + (0, a 2 ) so g(a 1 , a 2 ) = f 1 (a 1 ) + f 2 (a 2 ) = f (a 1 , a 2 ).
- I.7: 5. The point here is to use the construction given, and show that a family of maps fi : Ai → B de nes a unique map out of the disjoint union of the Ai.
- I.7: 6. I won’t verify the universal property here, just describe the construc- tion. (a) One can use the ordinary product (in S); one just has to take the basepoint of ∏
α∈I
Xα
to be the element that is xα 0 in the αth coordinate, where xα 0 is the basepoint of Xα. Note that the axiom of choice does not come into the existence of this element of
Xα since the Xα came with basepoints already assigned. (b) To make a coproduct, one takes the disjoint union of all the Xα except that one identi es all xα 0 to a single point (which will be the basepoint of the coproduct).
- X.1: 2. (a) There are lots of examples. The problem has to do with composition in Im T. Let's make a category C with two objects, A, B. hom(B, A) = hom(A, B) = ∅. hom(A, A) = { (^1) A, f } where f ◦ f = 1A, hom(B, B) = { (^1) B , g} where g ◦ g = 1B. This satis es the axioms of a category. Now let T (A) = T (B) = Z where T take values in S. Let T (f ) be the function ·(−1), and T (g) be the function that sends even integers to their negatives, and odd integers to themselves. Clearly T (f )◦T (g) is not in ImT , so ImT is not closed under composition, hence is not a category. 1
(b) If T is injective on objects, we need to check that Im T contains the identity and is closed under composition. If M is an object of ImT then M = T (A), so by the de nition of a functor, (^1) M = T (1A). If f : M → N , g : N → P are in ImT then M = T (A), N = T (B), P = T (C) uniquely, so f = T (α : A → B), g = T (β : B → C), so by de nition of a functor, g ◦ f = T (β) ◦ T (α) = T (β ◦ α), and hence is in ImT.
- X.1: 4a: The bijection is (a, b) 7 → (b, a). To check that this is natural, we are examining a map from P (A, B) = A × B to Q(A, B) = B × A. We call this map ηA,B : P (A, B) → Q(A, B). Naturality is commutativity of the diagram P (A, B) −−−−→ η(A,B)
Q(A, B)
yP^ (f,g)
yQ(f,g)
P (A′, B′) −−−−−−→ η(nA′,B′)
Q(A′, B′)
where f : A → A′, g : B → B′^ We calculate:
(1) (Q(f, g) ◦ ηA,B )(a, b) = Q(f, g)(b, a) = (g(b), f (a)) =
η(A′,B′)(f (a), g(b)) = (η(A′,B′) ◦ P (f, g))(a, b) which is as required. The associativity is similar. The two functors are (A, B, C) 7 → A × (B × C), and (A, B, C) 7 → (A × B) × C and the natural transformation takes (a, (b, c)) to ((a, b), c).
- X.1, 8b: Since Z is the free group on one generator, group homomorphism from Z to G are in one-one correspondence with elements of G. So the forgetful functor G → S is G 7 → hom(Z, G).
- I.1, 2: The unit is the map that take all of S to e ∈ G. f −^1 is the map s 7 → −f (s) (by −f (s) I mean the element in G inverse to f (s)). It is clear that inverse elements exist, and that the aforementioned identity element functions as the identity. To check associativity:
(2) (f + g) + h = (f + g)(s) + h(s) = (f (s) + g(s)) + h(s)
= f (s) + (g(s) + h(s)) = f + (g + h). On the third equality, we used associativity in G.
- I.1, 5: Normally one thinks of the symmetric group as the permutations of { 0 ,... , n − 1 }. For the purposes of this exercise it is helpful to think of the number of set isomorphisms between two different sets with n elements, say {a 1 ,... , an} and {b 1 ,... , bn} How many such isomorphism are there?
Now consider our polygon Xn, and position centered at (0, 0) as described, and with a vertex at (0, 1). Order the vertices counterclockwise so that the zeroth one is (0, 1). Clearly αn^ = e, the identity map, and the lower powers of α are all distinct since αi^ takes the zeroth vertex to the ith vertex. Note that they also preserve the counterclockwise order of the vertices, so αi^ takes the 1 st vertex to the (i + 1)st one. Also the elements αir, i = 0,... , n − 1 are all distinct from each other since they take the zeroth vertex to the ith one. They are also distinct from the αi^ since they simultaneously take the 1 st vertex to the (i − 1)st one. We need to show the existence of inverses and that this set is closed under multiplication.
(αir)−^1 = rαn−i
which must be in that set if it is closed under multiplication! To see that it is, note that αr = rα−^1 = rαn−^1. This can be checked by doing the matrix multiplication if the geometry is insuciently convincing. It follows that rαkrαj^ = rrαk(n−1)αj
so that the product of two things in our set is also in our set.
