# (Verification) If $f$ and $g$ are bijective, then so is $g \circ f$.

I made a post earlier just focusing on the injective case but now I've extended this to bijectivity:

Injectivity:

By injectivity (this isn't neccesary to write but I put it here for reference):

$$g(x)=g(x')\implies x=x'$$

$$f(x)=f(x')\implies x=x'$$

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

Therefore $$g\circ f$$ is injective.

Surjectivity:

Here is where I get confused, I usually get lost when it comes to surjectivity.

Suppose $$f: X \rightarrow Y, g:Y\rightarrow Z$$

For $$f(x)$$: $$\forall y\in Y, \exists x\in X:f(x)=y$$

For $$g(y)$$: $$\forall z\in Z, \exists y\in Y: g(y)=z$$

For $$g(f(x))$$: $$\forall z\in Z, \exists x:\exists y:g(y)=z$$. In other words, for any $$z$$, there is a $$y$$ that satisfies $$g(y)=z$$ since for any $$y$$ there is an $$x$$ that satisfies $$f(x)=y$$, thusly, surjectivity has occured.

Therefore $$g \circ f$$ is surjective, and thus it is bijective.

• Why do you insist on using logic symbols? You do it incorrectly and confuse yourself. – JCAA Jul 26 at 7:15
• @JCAA Where do I use it incorrectly? Honest question because I've always used them but I never questioned it. No one is there to correct me as I am doing self-study – Simplex1 Jul 26 at 7:16
• @Eevee Trainer Ah. That identity was later in the question so I couldn't use it. – Simplex1 Jul 26 at 7:18
• What you say about surjective seems correct, what confuses you? – Mikael Helin Jul 26 at 7:19
• @MikaelHelin I usually get questions on surjectivity wrong. I was wondering if I used the elements of the sets X, Y, Z correctly. – Simplex1 Jul 26 at 7:21

Deep Breath: And do it in the proper order.

For every $$z\in Z$$ we need to show there is an $$x\in X$$ so that $$g(f(x)) = z$$.

So we start in $$Z$$ and pull our way eventually to $$X$$ but go through $$Y$$ as the intermediate step.

So to go from $$z \in Z$$ what is it that gets up onto $$Z$$?

All this is me asking leading question to get you to note that we use $$g:Y\to Z$$ and so we start with $$g$$ being surjective. We do not start with $$f$$.

Now it's every bit as easy as they injective part!

$$g:Y \to Z$$ is surjective so for every $$z\in \mathbb Z$$ there is (at least) one $$y \in \mathbb Y$$ so that $$g(y) =z$$.

And $$f: X\to Y$$ is surjective so for the $$y$$ we used above (the specific $$y$$ so that $$g(y) = z$$, and not a generic $$y$$ in general) there is (at least) one $$x\in X$$ so that $$f(x) = y$$.

So for that $$x$$ we have $$g(f(x)) = g(y) = z$$ and for any $$z$$ we found at least one $$x \in X$$ so that $$g\circ f(x) = z$$.

So $$g \circ f$$ is surjective.

• Why can’t we start with $f$? The question states that it is injective. – Simplex1 Jul 26 at 21:31
• Because .... you are trying to make an argument... Every step must follow from the previos and lead to the next. If you start with $f$ you have for every $y$ there is an $x$ so that $f(x) = y$. But what the eff is $y$??? What's $y$ got to do with anything? It's just a value we pulled out of our ear. How is $y$ going to give us that $g(f(x))= z$ unless we know that $g(y) = z$? So we have to figure out that for $z \in Z$ there is a $y\in Y$ so that $g(y)=z$ first. If we do $g$ first we are fine butt if we do $f$ first we jumped into a lake with a pair of oars without getting a boat first. – fleablood Jul 26 at 21:42

"$$\forall z∈Z,\exists x:∃y:g(y)=z.$$"

This part isn't right because $$g$$ being surjective means that for all $$z \in Z$$ there exists $$y \in Y$$ such that $$g(y)=z$$. It has nothing to do with $$x \in X$$.

"In other words, for any z, there is a y that satisfies g(y)=z since for any y there is an x that satisfies f(x)=y."

The word "since" isn't right. Again $$X$$ and $$f$$ are totally irrelevant to $$g$$ being surjective. It is true that there is a $$y$$ such that $$g(y)=z$$. There is also an $$x \in X$$ that satisfies $$f(x)=y$$. Therefore $$(g \circ f)(x)=z$$, so $$g \circ f$$ is surjective.