2

I know if $g(f(x))$ is injective, then $f$ is injective and I have no problems proving this, but I also know g is not injective, but my following wrong proof suggests that $g$ is injective.

Proof: If $g(f(x))$ is injective, then if $g(f(x_1))=g(f(x_2))$, then $x_1=x_2$, so $f(x_1)=f(x_2)$ and $y_1=y_2$ for some $y_1$ and $y_2$. Since $g(f(x_1))=g(f(x_2))$ and $f(x_1)=f(x_2)$, so $g(y_1)=g(y_2)$ implies $y_1=y_2$, so $g$ is injective. Can anyone point out where my proof is wrong?

vitamin d
  • 5,783
Chihao Yu
  • 33
  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos May 16 '21 at 09:38
  • 2
    To prove that$g$ is injective you have to start with $g(y_1)=g(y_2)$ (and prove that $y_1=y_2$) not with $g(f(x_1))=g(f(x_2))$ – Kavi Rama Murthy May 16 '21 at 09:38
  • 1
    You can get a good intuition about this with drawing a Venn Diagram. if the composition is injective, you can only tell about $g(f(A))$ where A is the domain of the first function f. You can not figure out anything about the "rest" of the domain of the function g. – Idan Tank May 16 '21 at 09:41
  • But if I need to prove that if g(f(x)) is surjective, then g is surjective. One proof is:for every z in Z there exists x in X such that g(f(x))=z, and replace f(x) with y I can get g(y)=z, thus for every z, there exists y in Y. Why can I start with similar thing here? – Chihao Yu May 16 '21 at 09:49
  • Here's a counterexample: Take $f:{}\rightarrow{0,1}$ sending $\mapsto 0$ and $g$ the unique map ${0,1}\rightarrow{*}$. Your proof works only if every $y$ is $=f(x)$, for some $x$, so you only proved that $g$ is 1-1 in $f(A)$ – Alessandro May 16 '21 at 10:00
  • Take any injective function $f$ whose range is properly contained in the domain of $g.$ For instance let us take $f = \text {id}_{[0,1]}$ and $g : [-1,1] \longrightarrow \Bbb R$ defined by $g(x) = |x|,$ for all $x \in [-1,1].$ – Anacardium May 16 '21 at 10:07
  • Yeah, but as I said why can this proof be correct: if I need to prove that if g(f(x)) is surjective, then g is surjective. One proof is:for every z in Z there exists x in X such that g(f(x))=z, and replace f(x) with y I can get g(y)=z, thus for every z, there exists y in Y. Why can I start with similar thing(starting with g(f(x))=z,but not similar way for the wrong proof ) here? – Chihao Yu May 16 '21 at 12:22

0 Answers0