# Surjectivity and composition in functions

Sorry for this pretty dumb question, but I couldn't find any answer to it.

Supposing we have a set $$A$$ and functions $$f,g:A\to A$$. Prove that if $$f$$ isn't surjective then $$f \circ g$$ isn't surjective.

Basically as I think, In order for a composite function to be surjective' both the functions that are getting composited (In this case functions $$f$$ and $$g$$) should be surjective.

Is that right or wrong? And how can I prove that?

EDIT: Question is solved, Check my answer.

## 3 Answers

The statement is equivalent to $$f\circ g$$ surjective implies $$f$$ surjective, which is rather clear since : Let $$y\in A$$, there is $$x\in A$$ s.t. $$y=f(g(x))$$. In particular, if $$u=g(x)$$, then $$y=f(u)$$. We proved that $$\forall y\in A, \exists u\in A: y=f(u),$$ and thus surjectivity of $$f$$.

• Hey, I am sorry, but I have learned only the basics of group theory and functions, therefore I don't really know what these symbols mean: "∀ and ∃". Commented Aug 26, 2020 at 9:04
• $\forall$ means : for all and $\exists$ means : there exist.
– Surb
Commented Aug 26, 2020 at 9:07
• Thank you very much for your response. I have learned something new :) Commented Aug 26, 2020 at 9:10

Since $$f$$ is not surjective then

$$\exists y\in A\quad \forall x\in A\quad f(x)\neq y$$

then

$$\forall x\in A\quad z=g(x)\in A\quad f(g(x))=f(z)\neq y$$

therefore $$f \circ g$$ is not surjective.

It is given that $$f$$ isn't surjective. Because of that, There is $$y \in A$$ in such way that for each $$x\in A$$ exists $$f(x)\neq y$$.

So if, $$f \circ g$$ isn't surjective, then for the same element $$y$$ from beforehand, there is an origin in relation to the function $$f \circ g$$.

So there exists $$t\in A$$ in such way that $$f(g(t))=y$$.

But then, if we mark in $$x$$ the element $$g(t)$$ we will receive that $$f(x)=y$$ which is the opposite for what we have said at first.