# Homework Help: Function Injectivity Subjectivity Proof

1. Sep 17, 2016

### Austin Chang

1. The problem statement, all variables and given/known data
Proof Let F:A→B and g: B→C be functions. Suppose that g°f is injective. Prove that f is injective.

2. Relevant equations

3. The attempt at a solution
Let x,y ∈ A, and suppose g°f (x) = g°f(y) and x = y. Suppose g: B→C was not injective. then f(g(x)), if g(x) is some element within B, will be at least two numbers which would no longer make equation injective. If g:B→C was injective then f(c) = f(c') iff c = c'. Suppose f:A→B is not injective. Therefore if f(a') = f(a), a' = a does not need to be equal.
Therefore g°f will no longer be injective because a certain element within A can equal to multiple elements within C. Therefore F: A→B must be injective.

Tell me why my reasoning does not hold. My teacher sent me this reply. I don't know why my logic doesn't work.

"This is *way* too complicated. So much so, that it's not worth me checking whether your reasoning
holds in the end (although, I did notice some problems with that, too.) Follow the template! To show f is injective, let a, a' be in A and suppose f(a) = f(a'). Your job is to use the hypotheses to show that a = a'. It should not take more than a few short sentences."

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2. Sep 17, 2016

### SammyS

Staff Emeritus
Please give us the template that your teacher requested you to follow.

Your proof as it stands has many problems. It looks a little different than the hand written version you posted in an image, but that image is rather difficult to read. Both versions have problems.

3. Sep 17, 2016

### Austin Chang

Theorem 6 The function f : A → B is injective. Proof. Let x, y ∈ A, and suppose that f(x) = f(y). Then blah, blah, blah. It follows that x = y. Hence, f is injective.
Thanks

4. Sep 17, 2016

### SammyS

Staff Emeritus
Interesting template.

Here's one example of a difficulty with your proof. You state:
Suppose g: B→C ... if g(x) is some element within B ...​

For g: B→C, g(x) implies that x ∈ B and g(x) ∈ C . g(x) in general may not be anything like any of the elements found in set C.

5. Sep 17, 2016

### Austin Chang

So what would u say I should do? Do I just follow the format or do I just try to be more careful with what I say?

6. Sep 17, 2016

### Math_QED

Just follow the template. Injectivity is the fun part of these proofs.

Take $a, a' \in A$. Suppose $f(a) = f(a')$. Now try to deduce that $a = a'$. It really is a few sentences to write down as your teacher said.

Surjectivity is often a lot harder. Do you need to show something about surjectivity too (it's in the title of the thread)?

7. Sep 17, 2016

### SammyS

Staff Emeritus
Actually the title mentions subjectivity, not surjectivity .

Last edited: Sep 17, 2016
8. Sep 17, 2016

### SammyS

Staff Emeritus
@Austin Chang ,
In the OP, you inquired about shortcomings in you reasoning/logic.

You begin with:
Let x,y ∈ A, and suppose g°f (x) = g°f(y) and x = y.​
The first part of that statement OK logically. But the fact is that you are given that (g°f) is injective. This implies that x = y. This is not much use in showing that f injective.

Now if you join the first part of the statement with the part in red, you are assuming that x = y. If x = y then you have g°f (x) = g°f(y) simply from the definition of (g°f) being a well-defined function. In this case, g°f (x) = g°f(y) doesn't depend upon (g°f) being injective at all.

The you state:
Suppose g: B→C was not injective.​
It turns out that g does not need to be injective for (g°f) to be injective. You simply need g to be a function.

Continuing:
... then f(g(x)), if g(x) is some element within B, will be at least two numbers which would no longer make equation injective.​
Now we have problem heaped upon problem.
A big problem here is that you take elements from the wrong sets as inputs to functions and then assign the outputs (the images) to the wrong sets.
g(x) would be in set C, if x were in set B, but you already assigned x as being from set A. If g is a function, the g(b) never has two different outcomes. You also mention f(g(x)). For this to make sense, x needs to be in B, which then puts g(x) in C, but to use this in function f, you would need g(x) to be in A and the f(g(x)) would be in B. By the way, there is no need to consider f(g( )) at all, even if it existed. Finally, some terminology: An equation is not a function and thus is never injective.

No need to go on, is there ?

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