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

[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!

 

[sunjin09]

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

 

[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"

 

[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...

 

[sunjin09]

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"

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

 

[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.

 

[catherinenanc]

Does this apply to the problem?

 

[AlexChandler]

Does this apply to the problem?

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