4

STEVE FISK

Lemma 3

If

A and B are 2-path connected, then a map f:A#B ,.... C#D factors in one of these

ways:

(a) 7#6: A#B,.... C#D; 7:A,.... C; &B,.... D

(b)

"!#

6 : A#B ,.... C#D ; 7:A ,.... D ; &B ,.... C

(c) A#B,.... C,.... C#D

(d) A#B,.... D,.... C#D

Proof

By 2-path connectivity, it suffices to assume that A= B = fl.. Assume that f(lj) =

c#g0) for g: fl.,.... D and f(i,l) = h(i)#d for h :fl._. C. Note that h(l)=c and g(l)=d.

Either all first or second coordinates of f(ll.#j) are equal, so it is either c#/

1

(.i)

or

whO). Similarly, f(i#ll.) is either h(i)#u or /

2

(

i)#d. Since h(l) = c, h(i)

=f.

c for i

=f.

1.

Consequently, f(ij) is either c#d or h(i)#gO).

If

any f(ij) is h(i)#gO), then they all

are and we have case (c).

If

none are then c#d occurs twice in the image of some sim-

plex, and the map would not be non -degenerate.

By the above, we may assume that all simplices of fl.#

fl.

map to simplices of the

form u#p.

If

so, then f(ll.#j) = /

3

#d and f0#1l.) = /

4

#d for all i and j. Write f(ij) =

H(iJ)#d. His a map fl.# fl.,.... C, and so we have case d.

0

Corollary 4

If

A and B are 2-path connected, then Hom(A#B,C#D) = Hom(A#B,C)#D

U

Hom(A#B,D)#C

U

Hom(A,C)#Hom(B,D)

U

Hom(A,D)#Hom(B,C)

it is worthwhile to note that since A#B

is

not 2-path connected, we get two more

terms than afforded by Lemma 2.