Then ( g i : X → M i ) i is a fork if and only if g i ∉ ∑ j ≠ i Hom ( M j, M i ) g j for all i ∈ I. First, assume that there is some i with g i ∈ ∑ j ≠ i Hom ( M j, M i ) g j, say there is the subset { 1, 2, ⋯, t } ⊆ I such that g 1 ∈ ∑ j = 2 t Hom ( M j, M 1 ) g j.
"I was a fork lift driver," he said, glancing at his watch through welling tears.
Then the family ( u i : U → M i ) i ∈ I is a fork.
Scott Guthrie: I wouldn't have said it — I wouldn't say it's a fork.
"It was really hard, I'm quite optimistic about it now but at the time I kind of grieved - it was a fork in the road", she explained.
Then there exists for every index i ∈ I a simple submodule S i of M i, say with inclusion map u i : S i → M i, such that for every simple module S and I ( S ) = { i ∈ I ∣ S i = S } the family ( u i : S → M i ) i ∈ I ( S ) is a fork.
In order to see that ( u i : S → M i ) i ∈ I ( S ) is a fork, we have to show the following: For any finite subset J of I, say J = { 1, 2, ⋯, t }, the map u J : ( u i ) i : S → ⨁ i = 1 t M i is left minimal.
I am no longer the same self I was when I received the Happi-Fork; I am a policeman's girlfriend.
