# 1-1 and onto functions

Lee33

## Homework Statement

Let f: A→B be a surjection and g:B→C be such that g∘f is an injection. Prove that both f and g are injections

## The Attempt at a Solution

Suppose a = a' then g(f(a)) = g(f(a')) since g∘f is 1-1 we have that f(a) = f(a') hence f is 1-1.

Is this correct?

## Answers and Replies

Staff Emeritus
Gold Member
2021 Award
You assumed that a=a'and proved that f(a) = f(a'), which is simply showing that f is a function, not anything to do with whether f is 1-1. Furthermore, you went from g(f(a)) = g(f(a')) to f(a) = f(a') which seems to assume that g is injective already.

You never used the fact that f is a surjection, which should also tip you off that you missed something.

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Lee33
Let me redo.

Suppose a,a' ∈ A and that f(a) = f(a') then g(f(a)) = g(f(a')) since g∘f is an injection then we have a = a' therefore f is injective.

Is this correct?

Staff Emeritus
Gold Member
2021 Award
That looks better. Any ideas on what to do for g?

Lee33
Thanks! For g:

Let b, b' ∈ B. Then since f is onto there exists an a, a' ∈ A such that f(a) = b and f(a') = b'. Suppose g(b)=g(b') then we get g(b) = g(f(a)) = g(f(a'))= g(b') since g∘f is 1-1 then a = a' hence f(a) = f(a') therefore b = b' which means g is 1-1.

Is this correct?

Staff Emeritus
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