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Context

The problem I am doing is:

If $g \circ f$ is injective, must $g$ be injective?

I want to show, without an exampleu using finite sets, and by using characters like $\forall$ and $\exists$ that if $g \circ f$ is injective, $g$ may or may not be injective.

Why this post is different

I know how to do it, I know there are solutions on this site already. I am not asking how to solve it, I am asking about a specific phrasing of a proof of this.

Defining injective

I am defining injective as, \begin{align} \text{If $\psi:A\rightarrow B$ is “injective'', then }\left( \forall a_1,a_2\in A \right), (\psi(a_1)=\psi(a_2)) \wedge (a_1=a_2). \end{align}

Now I want to show that if $g \circ f$ is injective, it does not matter if $g$ is injective.

Question

Is the only way to explicitly show that $g$ may be injective is to assume it is not and show that $g \circ f$ prevails to be injective?

Example

Suppose that $\text{ }f: A\rightarrow B$ and $g: B \rightarrow C$ are functions and $f$ is injective. Assume $g$ is not injective: $$ \exists b_1, b_2 \in B: (g(b_1) = g(b_2)) \wedge (b_1 \neq b_2) $$

If $\forall c \in A: \text{ } f(c) \neq b_1, b_2$, then $g\circ f$ is injective.

Is there anything I can add to this example to make it more clear that $g$ can be injective or not?

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    I would give two concrete examples, one in which $g$ is injective and one in which $g$ is not injective, but in both cases $g\circ f$ is injective. For instance, let $f:\mathbb{R}\to \mathbb{R}$ be $f(x)=e^x$, and then set $g_1:\mathbb{R}\to\mathbb{R}$, $g_2:\mathbb{R}\to\mathbb{R}$, with $g_1(x)=x^2$ and $g_2(x)=x$. – 2017-02-09
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    @ThomasGrubb The OP is not asking a math problem. The OP is asking a notation problem. The OP (presuably) knows how to solve this and is not interested in any actual solution. The OP simply wants to know how to write "can be but doesn't need to be" symbolically. – 2017-02-09
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    @fleablood Ok, I guess I skipped over his desire to do it symbolically. However, for me, two examples such as above both constitute a proof of what he is asking and are much clearer than trying to parse the symbols. – 2017-02-09
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    Actually rereading I'm not sure if he is asking for symbols or simply a proof without counter example. So sorry if it seemed like I was jumping down your throat. – 2017-02-09
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    Maybe you should look into a logic with more than 2 values. This logic is called "modal logic". See for example: https://en.wikipedia.org/wiki/Modal_logic – 2017-02-09
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    You say you want a proof without using an example and then end your post with an abstract example. It might be helpful for you to think about what you mean by example. – 2017-02-09
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    I am sorry, by example I meant using finite sets. – 2017-02-09
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    I do appreciate your comment ThomasGrubb. To be honest I am asking both ThomasGrubb's question and fleablood's first assumption on what I asked, it was not clear to me of the distinction at the time. – 2017-02-09
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    I misspoke again, by finite sets I mean like $A=\{1,2\}$, what would I call this in comparison to, $A \subset \mathbb N$ where A is finite? – 2017-02-09

2 Answers 2

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"P(x) does not necessarily imply Q(x)" is equivalent to "There is some x with P(x) and not Q(x)."

"P(x) does not rule out Q(x)" is equivalent to "There is some y with P(y) and not Q(y)."

Combining these should give you what you want.

That said, mathematical notation isn't a programming language. It's meant to be read by humans. So don't translate things into complicated symbolic expressions without good reason.

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To say that $g$ need not be injective if $g\circ f$ is injective is the same as saying that at least one counterexample exists. I would simply exhibit the simplest possible counterexample: \begin{align} \operatorname{domain}(f) & = \{1\} & & & \operatorname{domain}(g) & = \{1,2\} \\[10pt] f(1) & = 1 & & & g(1) & = 1 \\ & & & & g(2) & = 1 \end{align} For the functions above, $g\circ f$ is injective, but $g$ is not injective.

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    But... $g$ *is* injective. – 2017-02-09
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    Thank you Michael, my question was not clear and I edited it to make it more clear. I do not want to find a solution using finite and explicitly defined sets. – 2017-02-09
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    TYPO, I meant $g(2) = 1. \qquad$ – 2017-02-09