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I would like to know whether a function I have thought up is injective or not. For reference; the definition of a function that is injective may be expressed as ($\ref{1}$) https://en.wikipedia.org/wiki/Injective_function

$(\forall a,b \in X)([f(a)=f(b)]\rightarrow [a=b])\label{1}\tag{1}$

I have constructed (I think) a very weird injective function that only has 1 element in its domain and co-domain but accepts multiple inputs at once.

Imagine a function which has domain {1} and co-domain {1}. It is injective because it is one to one. Now Imagine that the functions domain is the truth value of an expression in propositional logic as opposed to a number.

We may write a domain of {$A\lor\lnot A$} and co-domain {1} for this case.

Next we can imagine writing {$A\lor\lnot A=B\lor\lnot B$} and co-domain {1} The function is this case will accept inputs A and B despite their ability to be completely different things. Even different numbers So far as I can tell: it does not deviate from the definition of an injective function. Which is for all a and b in X, f(a) = f(b) implies a = b ($\ref{1}$)

Is such a function injective? Thanks in advance.

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You're equivocating between "truth value of an expression" and "expression".

When viewed as a function from {expressions with truth value 1} to $\{1\}$, it's massively non-injective. When viewed as a function from {truth value 1} to $\{1\}$, it's injective.

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    I am unsure of the difference between an expression with truth value of 1 and a truth value of 1. Why are they not the same? Also, just to clarify, I am not re-purposing the value of 1 to mean true. I am just creating a function that always outputs the same number so long as the input is valid. – 2017-02-12
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    I'm sure you agree that "A or not A" is not literally $1$, though it has a truth value of $1$. When the function is viewed from the set {expressions whose truth value is 1}, it's not injective. When viewed from the set {literal truth value 1} it's injective, because that set has cardinality $1$. Note that "A or not A" is in the first set but not the second, and 1 is in the second set but not the first (since truth values are not formulae). – 2017-02-12
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    Note that in my original question {A∨¬A} was a set that contained a single element. An example of a set which does not might be {A∨¬A , B∨¬B}. Similarly, the set {A∨¬A=B∨¬BA∨¬A=B∨¬B} also contained only a single element. This means they both have cardinality 1. – 2017-02-12
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    I think I don't understand your question, then. *Any* function from a one-point set to a one-point set is injective, though they may be completely different functions. There's nothing special about $A \vee \neg A$ or $B \vee \neg B$ here; the function from {False} to {1} is injective just the same. What do you mean by "their ability to do completely different things" in your original post? – 2017-02-12
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    By that I meant that multiple different inputs could be used for this function Yet it would output the same result. In English one could say that it was giving the same output for deferent inputs; which is not within the intent of what an injective function is supposed to do. – 2017-02-12
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    @user400188 you mix up the injective $f:\{1\}\to\{1\}$ with the non-injective $f\circ g:E\to\{1\}$ where $E$ denotes some class of expressions having truth value $1$ and $g:E\to\{1\}$. – 2017-02-12
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    @drhab that seems to answer the question. I had suspected that the function could be defined that way (using function composition) but I was not advanced enough to know that it was the only way it could be defined. Thank you – 2017-02-12