Prove that if $f:A\rightarrow B$ and $g:B\rightarrow A$ are such functions, that $gf = I_A$, then $f$ is 1-1 and $g$ is onto.
Knowing that $gf = I_A \equiv gf : A\rightarrow A$, I had an idea to present $gf$ function as
$$gf(x) = x$$
Then for any element $x\in A$, function $gf$ is true.
But what should I do next in order to prove that $f$ is 1-1 and $g$ is onto?
Prove that if $f:A\rightarrow B$ and $g:B\rightarrow A$ are such functions, that $gf = I_A$, then $f$ is 1-1 and $g$ is onto.
elementary-set-theory
Related Solutions
A picture like this one may help a bit:
The black arrows show the function $f:A\to B$; the colored blobs show elements of $\wp(A)$ on the left and their images under $g$ in $\wp(B)$ on the right. The colored arrows show the action of $g$ on these sets.
Added: I’ll do one direction and leave the other to you. Suppose that $f$ maps $A$ onto $B$. To show that $g$ is onto, I must show that for each $Y\subseteq B$ there is some $X\subseteq A$ such that $g(X)=Y$, so let $Y$ be an arbitrary subset of $B$. Let $X=\{x\in A:f(x)\in Y\}$. (In other words, $X=f^{-1}[Y]$.) I claim that $g(X)=Y$.
To see this, recall that $g(X)=\{f(x):x\in X\}$; clearly $f(x)\in Y$ for each $x\in X$, since we chose $X$ to make that true, so $g(X)\subseteq Y$. On the other hand, the map $f$ is onto, so for each $y\in Y$ there is an $x\in A$ such that $f(x)=y$. Clearly this $x_y\in X$, so for each $y\in Y$ there is an $x\in X$ such that $f(x)=y$. Thus, $g(X)\supseteq Y$. Combining the inclusions, we see that $g(X)=Y$, so $Y$ is in the range of $g$. $Y$ was an arbitrary subset of $B$, so every subset of $B$ is in the range of $g$, and therefore $g$ must map $\wp(A)$ onto $\wp(B)$.
Now see if you can show that if $g$ is onto, so is $f$.
You wanted $h:A\to C$, but when you say (for $a\in A$) that you're letting $h(a)=f(a)$, you're defining $h:A\to B$. There are other, similar errors, which amount to failing to keep track of your (co)domain(s).
You're quite correct that we can use $f:A\to B$ and $g:B\to C$ to define $h$, though. What can you say about $g\circ f$, given what you know of $f$ and $g$?
Related Question
- [Math] Suppose $f: A\rightarrow B,g:B\rightarrow C, h:B\rightarrow C$. Pro that If $f$ is onto $B$ and $g\circ f$= $h\circ f$, then $g=h$
- [Math] Prove that if $g \circ f$ is onto and $g$ is one-to-one, then $f$ is onto
- [Math] $f:A\to B$ and $g:B\to A$ two functions such that $f\circ g\circ f$ is bijective. Then so are $f, g$
Best Answer
For every $x\in A$, $x=gf(x)$, so $g$ in onto.
For every $x,y\in A$, if $f(x)=f(y)$ then $x=gf(x)=gf(y)=y$, so $f$ is 1-1.