1

Show that if $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are injective, the $g \circ f$ is injective.

Firstly, we have $g \circ f:X \rightarrow Z $. If $g \circ f$ is injective, suppose $(g \circ f)(x) = (g \circ f)(x')$ where $ x, x' \in X$. Then, $(g \circ f(x)) = (g \circ f(x'))$. From this it can be concluded $g(y) = g(y')$ where $y, y' \in Y$ . Since $g$ is injective we have $y = y'$. Hence, $g \circ f$ must also be injective. $\square $

Is this proof acceptable?

Retty
  • 487

4 Answers4

0

You are making a mistake. The proper way to do it is as follows:

Suppose $g\circ f(x)=g\circ f(x')$. Then, by definition, $g(f(x))=g(f(x'))$. Since $g$ is injective, we have that $f(x)=f(x')$. Since $f$ is injective, the latter implies that $x=x'$, but this is what we needed to show, so we're done.

Mathematician 42
  • 12,591
  • Oh right, that makes more sense. Thanks @Mathematician 42! – Retty Mar 17 '17 at 08:53
  • In reference to my comment under the question: if $y$ and $y'$ are the results of $f(x)$ and $f(x')$ respectively, then it is not equivalent to saying $f(x) = f(x')$? – Retty Mar 17 '17 at 08:58
  • Yeah, but you still need the injectivity of $f$ to conclude that $x=x'$ from $y=y'$. And you certainly need to remove the assumption that $g\circ f$ is injective as that is what you want to show. – Mathematician 42 Mar 17 '17 at 09:03
  • Thanks @Mathematician 42, it's certainly a lot clearer now. – Retty Mar 17 '17 at 09:05
0

The clause "if $g\circ f$ is injective" is wrong; it is what to be proved. Take out this clause and begin your proof with just "suppose $g\circ f(x) = g\circ f (x')...$". Then it would be okay to read.

If you are after something more meticulous, then I would say you should familiarize yourself with the use of math quantifiers (for all, there is some, there is exactly one, ...).

Yes
  • 20,719
  • @Retty No, and two other answers explain why it is not OK. – 5xum Mar 17 '17 at 08:51
  • @5xum, ? We must have different definition of being okay. By okay I mean "I can guess what the author tried to say". But as I pointed out, an unfixable logical error is not okay. – Yes Mar 17 '17 at 08:53
  • My answer was just an answer to OP. Sorry for the confusion – 5xum Mar 17 '17 at 08:54
0

Your proof has multiple flaws.

First of all, you say "if $g\circ f$ is injective", and that's a very bad way to start a proof. You aren't allowed to suppose that $g\circ f$ is injective, you have to prove that it is.

Using your method, I can say "if I am a martian, then I am a martian", but that doesn't really prove anything.

Second, you use the expression $y$ and $y'$ without defining what $y$ is.

Third, you don't explain how $y=y'$ implies that $g\circ f$ is injective. What you have to prove is that $x=x'$!

5xum
  • 123,496
  • 6
  • 128
  • 204
0

Suppose there are $x,\ x' \in X$ such that $g(f(x)) = g(f(x'))$. By the injectivity of $g$, necessarily $f(x) = f(x')$ and by the injectivity of $f$, we get $x = x'$ like we wanted to. This means that the composition is injective.

qualcuno
  • 17,121