Problem Set - Functions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 22 July 2011

Functions

Let S, T and U be sets and let f:ST and g:TU be functions. Show that
(a)   if f and g are injective then gf is injective,
(b)   if f and g are surjective then gf is surjective, and
(c)   if f and g are bijective then gf is bijective.

Let f:ST be a function and let US. The image of U under f is the subset of T given by f(U)= {f(u) | uU}.

Let f:ST be a function. The image of f U under f is the is the subset of T given by imf= {f(s) | sS}. Note that imf=f(S).

Let f:ST be a function and let VT. The inverse image of V under f is the subset of S given by f-1(V) = {sS | f(s)V}.

Let f:ST be a function and let tT. The fiber of f over t is the subset of S given by f-1(t) = {sS | f(s)=t}.

Let f:ST be a function. Show that the set F= {f-1 (t) | tT} of fibers of the map f is a partition of S.

(a)   Let f:ST be a function. Define f:S imf s f(s) Show that the map f is well defined and surjective.
(b)   Let f:ST be a function and let F={ f-1(t) | timf} = { f-1(t) | tT} - be the set of nonempty fibers of the map f. Define f^: F T f-1(t) t Show that the map f^ is well defined and injective.
(c)   Let f:ST be a function and let F={ f-1(t) | timf} = { f-1(t) | tT} - be the set of nonempty fibers of the map f. Define f^: F imT f-1(t) t Show that the map f^ is well defined and bijective.

Let S be a set. The power set of S, 2S, is the set of all subsets of S.

Let S be a set and let {0,1}S be the set of all functions f:S{0,1}. Given a subset TS define a function fT:S {0,1} by fT(s) = { 0, ifsT, 1, ifsT.

Show that φ: 2S {0,1}S T fT is a bijection.

Let :S×S S be an associtaive operation on a set S. An identity for is an element e S such that if sS then es =se=s.

Let e be an identity for an associative operation on a set S. Let sS. A left inverse for s is an element t S such that ts=e. A right inverse for s is an element tS such that st=e. An inverse for s is an element s-1S such that s-1s = ss-1 =e.

  1. Let be an operation on a set S. Show that if S contains an identity for then it is unique.
  2. Let e be an identity for an associative operation on a set S. Let sS. Show that if s has an inverse then it is unique.

  1. Let S and T be sets and let ιS and ιT be the identity maps on S and T, respectively. Show that for any function f:ST, ιTf=f, and fιS=f.
  2. Let f:ST be a function. Show that if an inverse function to f exists then it is unique. (Hint: The proof is very similar to the proof in Ex. 5b above.)

Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

page history