$f$ is bijective, show that $h(x)=\left(f(x); g(x)\right) \rm{\ is\ bijective\ } \iff G $ is Singleton

Question

Let E, F and G be three sets ($E\neq 0;F\neq 0,G\neq 0 ) $ Let $h$ defined by : $$\begin{align} h \ \colon\ E & \to F\times G\\ x & \mapsto h(x)=\biggl(f(x);g(x)\biggr). \end{align}$$ with $\begin{align} f \ \colon\ E & \to F \\ x & \mapsto f(x). \end{align}$ and $ \begin{align} g \ \colon\ E & \to G \\ x & \mapsto g(x). \end{align}$

Suppose that $f$ is bijective. show that :

$h \rm{\ is\ bijective\ } \iff G $ is Singleton

Indeed;

• Show that $$ h \rm{\ is\ bijective\ } \implies G \rm{\ is\ Singleton } $$

Proof by contradiction:

suppose that G contains at least two elements $(a,b)\in G^2$ such that $a\neq b$ Let $y\in F$ and since $f$ is bijective then $\exists ! x\in E $ such that $f(x)=y$, since $h$ is bijective and $(y,a)\in F\times G$ then $\exists ! x \in E$ such that $h(x)=(y;a)$ the same thing for $(y;b)\in F\times G$ then $\exists ! x \in E$ such that $h(x)=(y;b)$ thus $g(x)=a=b$ which is contradictory .

• Show that $$ G \rm{\ is\ Singleton } \implies h \rm{\ is\ bijective\ } $$ Let $G=\{a \}$ then $\forall x \in E,\quad h(x)=\left(f(x);a \right)$

• i'm stuck here how can i prove that h is bijective

Answer 1

In general a map $k:X\to Y$ is bijective if and only if it has an inverse . This is actually a map $l:Y\to X$ such that $l\circ k=\text{id}_X$ and $k\circ l=\text{id}_Y$.

This map $l$ is unique, bijective and is denoted as $k^{-1}$. Its (unique) inverse is the map $k$.

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Function $f:E\to F$ in your question is bijective, so has an inverse $f^{-1}:F\to E$.

Then the map $h:E\to F\times\{a\}$ prescribed by $x\mapsto\langle f(x),a\rangle$ has an inverse in the map $F\times\{a\}\to E$ that is prescribed by $\langle y,a\rangle\mapsto f^{-1}(y)$.

Answer 2

Let’s start by getting an intuition for what’s going on. Since $f$ is a bijection, $F$ is intuitively just $E$ by another name: for $x\in E$ we can think of $f(x)$ as just a sort of alias for $x$. Thus, the special case in which $E=F$ and $f$ is the identity function should be representative of all possibilities.

In that special case $h(x)=\langle x,g(x)\rangle$ for each $x\in E$. Clearly $h$ is injective: if $x_0\ne x_1$, then

$$h(x_0)=\langle x_0,g(x_0)\rangle\ne\langle x_1,g(x_1)\rangle=h(x_1)\;.$$

Now fix $x_0\in E$, suppose that there is some $y\in G\setminus\{g(x_0)\}$, and let $x\in E$ be arbitrary. Then $h(x)=\langle x,g(x)\rangle$, and either $x\ne x_0$, in which case $h(x)\ne\langle x_0,y\rangle$, or $x=x_0$, in which case $h(x)=\langle x_0,g(x_0)\rangle\ne\langle x_0,y\rangle$. Thus, $\langle x_0,y\rangle$ is not in the range of $h$, and $h$ is not surjective (and hence not bijective). On the other hand, if $G=\{g(x_0)\}$, then $g(x)=g(x_0)$ for all $x\in E$, and the function $h$ is simply $h(x)=\langle x,g(x_0)\rangle$ for all $x\in E$. But then for each $\langle x,y\rangle\in F\times G=E\times G$ we have $y=g(x_0)$ and hence $\langle x,y\rangle=\langle x,g(x_0)\rangle=h(x)$, so that $h$ is surjective (and hence a bijection).

Now see if you can make the relatively slight modification needed to replace the identity map with an arbitrary bijection $f:E\to F$. It may help to note that if $i:E\to E$ is the identity map, when I initially wrote that in the special case we have $h(x)=\langle x,g(x)\rangle$, I could just as well have written $h(x)=\langle i(x),g(x)\rangle$. If you go through and find where a plain $x$ or $x_0$ is being thought of as $i(x)$ or $i(x_0)$ — i.e., where it’s being thought of as a member of the codomain of $i$ rather than as a member of the domain — you can pretty much just replace $i$ by $f$.