Let U be a subset of X.12

  1. 1. The image part of the direct image with compact support f!(U) of U is the set f!,im(U) defined by
    f!,im(U)=deff!(U)Im(f)={yY | we have f1(y)U and f1(y)Ø}.
  2. 2. The complement part of the direct image with compact support f!(U) of U is the set f!,cp(U) defined by
    f!,cp(U)=deff!(U)(YIm(f))=YIm(f)={yY | we have f1(y)U and f1(y)=Ø}={yY | f1(y)=Ø}.


1Note that we have

f!(U)=f!,im(U)f!,cp(U),

as

f!(U)=f!(U)Y=f!(U)(Im(f)(YIm(f)))=(f!(U)Im(f))(f!(U)(YIm(f)))=deff!,im(U)f!,cp(U).

2In terms of the meet computation of f!(U) of Remark 2.6.3.1.3, namely

f!(χU)=xXf(x)=1(χU(x)),

we see that f!,im corresponds to meets indexed over nonempty sets, while f!,cp corresponds to meets indexed over the empty set.


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