I'm trying to prove that composition of injections is an injection. I want to know if this is a good proof:
Composition of injections is an injection.
Let $f:S_1\rightarrow S_2$ and $g:S_2\rightarrow S_3$ be injections by the definitions:
$1.~\forall x_1, x_2:x_1\ne x_2\Rightarrow f(x_1)\ne f(x_2)\\2.~\forall y_1, y_2:y_1\ne y_2\Rightarrow g(y_1)\ne g(y_2)$
then composition $g\circ f(x)$ is an injection.
We have to prove:
$\forall x_1, x_2:x_1\ne x_2\Rightarrow g\circ f(x_1)\ne g\circ f(x_2)$.
Let $x_1, x_2\in S_1$ and $x_1\ne x_2$, then by the first definition of injectivity of $f$, $f(x_1)\ne f(x_2)$. Let $f(x_1)=y_1$ and $f(x_2)=y_2$. It follows that $y_1\ne y_2$. Then by the second definition of injectivity of $g$, $g(y_1)\ne g(y_2)$.
Equations imply:$$g(y_1)\ne g(y_2)\\g[f(x_1)]\ne g[f(x_2)]\\g\circ f(x_1)\ne g\circ f(x_2).$$
Since we started with general $x_1$ and $x_2$ this is proved.
Or without the introduction of $y_1$ and $y_2$, so just "… By the second definition of injectivity: $f(x_1)\ne f(x_2)$ then $g[f(x_1)]\ne g[f(x_2)]$. Therefore $x_1\ne x_2$ imply $g\circ f(x_1)\ne g\circ f(x_2)$."
Best Answer
It's good, but proving injectivity is often easier with the contrapositive:
So, assume $f$ and $g$ are injective and suppose $$ g\circ f(x_1)=g\circ f(x_2). $$ This means $$ g(f(x_1))=g(f(x_2)) $$ that, by the injectivity of $g$, implies $$ f(x_1)=f(x_2). $$ Injectivity of $f$ allows us to argue that $$ x_1=x_2. $$
As you clearly see, these are the same steps as in your proof; but dealing with equality is psychologically easier than dealing with inequalities.