If g°f is one-to-one, must f and g both be one-to-

1. Mar 15, 2012

catherinenanc

1. If g°f (g composed with f, two functions) is one-to-one, must f and g both be one-to-one?

2. The answer is, no, only f has to be, but I just can't see why!

2. Mar 15, 2012

sunjin09

Obviously g needs to be one to one when restricted to Im(f) ...

3. Mar 15, 2012

catherinenanc

Apparently not! I first guessed that both f and g had to be one-to-one, because I could not draw a map otherwise, but the graded work sent back to me said "No! Only f has to be 1-1"

4. Mar 15, 2012

catherinenanc

Also, the next problem is: "Let f:A→B and g:B→C be maps such that g°f is injective. Prove that f must be injective."

So, twice he's said that it is f that must be injective. I just can't figure out why g doesn't have to be...

5. Mar 15, 2012

sunjin09

Your grader is right, now try to read my post again and complete the ... part, starting with a 'but'.

6. Mar 15, 2012

catherinenanc

Ok, I am seeing an example and a counterexample.

f={(1,1),(2,2)} and g={(1,1),(2,2),(3,2)}. Then f is one-to-one, g is not, and g°f={(1,1),(2,2)} is. Right? This works because the domain of g is not restricted to the image of f.

Also, f={(1,1),(2,2),(3,2)} and g={(1,1),(2,2)}. Then f is not one-to-one, g is, and g°f={(1,1),(2,2),(3,2)} is not.

7. Mar 15, 2012

catherinenanc

Does this apply to the problem?

8. Mar 15, 2012

AlexChandler

yes that is right, but the last part does not prove anything, only the first part is important