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


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.


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.

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    $\begingroup$ Why do you insist on using logic symbols? You do it incorrectly and confuse yourself. $\endgroup$ – JCAA Jul 26 at 7:15
  • $\begingroup$ @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 $\endgroup$ – Simplex1 Jul 26 at 7:16
  • $\begingroup$ @Eevee Trainer Ah. That identity was later in the question so I couldn't use it. $\endgroup$ – Simplex1 Jul 26 at 7:18
  • $\begingroup$ What you say about surjective seems correct, what confuses you? $\endgroup$ – Mikael Helin Jul 26 at 7:19
  • $\begingroup$ @MikaelHelin I usually get questions on surjectivity wrong. I was wondering if I used the elements of the sets X, Y, Z correctly. $\endgroup$ – 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.

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  • $\begingroup$ Why can’t we start with $f$? The question states that it is injective. $\endgroup$ – Simplex1 Jul 26 at 21:31
  • $\begingroup$ 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. $\endgroup$ – 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.

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