# Injectivity and Surjectivity of Composition Function

The question asks:

Let $$g : A \to B$$ and $$f : B \to C$$ be two functions.

Show that if $$f \circ g$$ is a bijection, then $$f$$ is a surjection and $$g$$ is an injection

I know how to prove if it's given that either $$f$$ or $$g$$ is either injective or surjective, but I am not quite sure how to approach this question since it's asking for both.

Also, how do I show that $$f \circ g$$ is a bijection?

## 3 Answers

First, we note that $$f \circ g$$ will be a map $$A \to C$$. If we assume that $$f \circ g$$ is bijective, we automatically have the following two observations:

1. Surjectivity: for each $$c \in C$$, there exists $$a \in A$$ such that $$(f \circ g)(a) = c$$.
2. Injectivity: if $$(f \circ g)(a) = (f \circ g)(a^\prime)$$, then we must have $$a = a^\prime$$.

All that is needed from here are the two properties above. I'll write out the injectivity part and leave the surjective part to you.

To check that $$g$$ is injective, we proceed directly using the definition. Let's assume that $$g(a) = g(a^\prime)$$ for two $$a,a^\prime \in A$$. Applying $$f$$ to both sides of this equation, we find that $$(f \circ g)(a) = f(g(a)) = f(g(a^\prime)) = (f \circ g)(a^\prime)$$ whence we must have $$a = a^\prime$$ (why? which property is used here?). Using 1. in a similar way, we can show that $$f$$ must be surjective as well.

Also, the hypothesis is that $$f \circ g$$ is bijective. Consequently, you do not need to prove that $$f \circ g$$ is bijective as this property is given!

Use the fact that $$f\circ g$$ is a surjection to prove that $$f$$ is a surjection, too. As a great big hint: $$(f\circ g)(x)=f\bigl(g(x)\bigr).$$

Use the fact that $$f\circ g$$ is an injection to prove that $$g$$ is an injection, too. As a great big hint: $$(f\circ g)(x)=f\bigl(g(x)\bigr)$$ and $$(f\circ g)(y)=f\bigl(g(y)\bigr).$$

• how about proving f◦ g is a bijection? – ph-quiett Apr 7 at 19:04
• That isn't what you've been asked to do. Rather, you're supposed to assume that $f\circ g$ is a bijection, and use that to prove facts about $f$ and $g.$ – Cameron Buie Apr 7 at 19:18

Hint:

Prove these more general assertions:

1. If $$f\circ g$$ is injective, then $$g$$ is injective (by contrapositive).
2. If $$f\circ g$$ is surjective, then $$f$$ is surjective (observe that $$(f\circ g)(A)\subset g(B)$$).
• There's really no need for contrapositive, since $f$ is a function. – Cameron Buie Apr 7 at 19:19
• Injectivity is really very short by contrapositive. – Bernard Apr 7 at 19:22
• True. The direct proof is shorter, though. – Cameron Buie Apr 7 at 19:25
• Shorter than this: ‘If $g$ is not injective, obviously $f\circ g$ can't be either.’? – Bernard Apr 7 at 19:40
• I would contend that use of the word "obviously" would be ill-advised at this level of mathematics, so wouldn't constitute an acceptable proof, unless followed up by "because [justification from the definition]." – Cameron Buie Apr 7 at 19:48