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.
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