Proof of Injectivity, Composition, and Restriction of Bijections

Throughout we use the defining property of a bijection $u:X\to Y$: for every $y\in Y$ there is exactly one $x\in X$ with $u(x)=y$.

Claim 1. Let $x,x'\in X$ satisfy $u(x)=u(x')$, and call this common value $y\in Y$. Then $x$ and $x'$ are both elements of $X$ mapped by $u$ to $y$. Since there is exactly one such element, $x=x'$.

Claim 2. Let $z\in Z$. Since $v$ is a bijection, there is exactly one $y\in Y$ with $v(y)=z$; since $u$ is a bijection, there is exactly one $x\in X$ with $u(x)=y$. We show that $x$ is the unique element of $X$ with $w(x)=z$. First, $w(x)=v(u(x))=v(y)=z$. Second, let $x'\in X$ satisfy $w(x')=z$, that is $v(u(x'))=z$. By uniqueness of the $v$-preimage of $z$ we get $u(x')=y$, and by uniqueness of the $u$-preimage of $y$ we get $x'=x$.

Claim 3. Write $u_A$ for the restriction of $u$ to $A$. By the definition of $u(A)$, the map $u_A$ takes values in $u(A)$. Let $y\in u(A)$. Then $y=u(a)$ for some $a\in A$, so $u_A(a)=y$. If $a'\in A$ also satisfies $u_A(a')=y$, then $u(a')=u(a)$, so $a'=a$ by Claim 1. Hence there is exactly one $a\in A$ with $u_A(a)=y$, and $u_A$ is a bijection from $A$ onto $u(A)$.