# Show that the composition function and its inverse are one to one and onto

Let both $f: A \mapsto B$ and $g: B \mapsto C$ be one-to-one and onto. Show that the composition $g \circ f : A \mapsto B$ and the inverse $f^{-1} : B \mapsto A$ are one-to-one and onto.

Here is what I have so far:

Show onto: By definition of composition

$$g \circ f (x) = g(f(x))$$

So for any $b \in B$, there exist an $a \in A$ such that $f(a)=b$, so $$f(a)=b$$ thus $g(b)= c$ for all $c\in C$ which to me is not correct.

Showing one to one: For all $a \in A$ there exist a $b \in B$ such that $f(a) = f(b)$ where $a=b$.

Can anyone point me in the right direction?

• @String I edited the post for readability, and that was one of the formatting changes I made as well. It will reflect my edits soon. – NoseKnowsAll Sep 15 '15 at 20:38
• math.stackexchange.com/questions/1364303/definition-of-composition-of-functions – R.N Sep 15 '15 at 20:44

What you have written seems pretty confused. Let's look at the "onto" portion given bijections (one-to-one and onto functions) $A \overset{f}{\to} B \overset{g}{\to} C$.

You say that

By definition of composition $g \circ f (x) = g(f(x))$ so for any $b \in B$ there exists an $a \in A$ such that $f(a) = b$ [...]

But this gives the impression that somehow composition implies the existence of such an $a \in A$. Such an $a \in A$ is guaranteed to exist, but it's because the function $f$ maps $A$ onto $B$.

For all $a \in A$ there exist a $b \in B$ such that $f(a)=f(b)$ where $a=b$

shows that you haven't really understood the definition of one-to-one functions. It's really an "If... Then..." statement; if we know that $f(x_1) = f(x_2)$ and that $f$ is one-to-one, then we know that $x_1 = x_2$ (otherwise, two distinct elements of the domain, $x_1$ and $x_2$, would get mapped to the same element).

You will absolutely not be able to prove things like this if you don't have a firm understanding of the definitions. It's also good practice, at first, to write down exactly what you need to prove: The assumptions, and the conclusion you're looking for. Justify each claim you make, and pay close attention to what you've written. For example...

I would argue that $f \circ g$ maps $A$ onto $C$ as follows.

We need to show that, given any $c \in C$, there exists some $a \in A$ so that $(g \circ f)(a) = g(f(a)) = c$.

Since $g$ maps $B$ onto $C$, we know that there exists some $b \in B$ so that $g(b) = c$. But then, since $f$ maps $A$ onto $B$, we can find some $a \in A$ so that $f(a) = b$ hence

$$g(f(a)) = g(b) = c,$$

as desired.

To show that $g \circ f$ is one-to-one, you'll pick $a_1, a_2 \in A$ and assume that $g(f(a_1)) = g(f(a_2))$. Your job will be to show that $a_1 = a_2$, and I'll leave that to you (Hint: Since $g$ is one-to-one, what can we say about $f(a_1)$ and $f(a_2)$?) .

• Thank you so very much i will finish it then post it, i was sort of confused in this problem. – user146269 Sep 15 '15 at 21:40
• @user146269 I completely understand; I just wanted to point out that certain things stuck me as odd, and hopefully you'll be able to see how. Reasoning about these sorts of things can be very difficult at first, but once you get your bearings, it usually gets much easier. – pjs36 Sep 15 '15 at 22:01
• Note that after the horizontal rule, I want to say "I would argue that $g \circ f$ maps $A$ onto $C$..." I had the composition backwards, but my actual work doesn't make that mistake. – pjs36 Sep 16 '15 at 16:08
• Yes, do not worry about that i know – user146269 Sep 16 '15 at 17:36

Remember:

For onto: ‘$g\circ f$ onto’ means every element in $C$ has a pre-image in $A$.

For one-to-one: it means different elements in $A$ have different images in $C$.