1. Sets

Let A,B, and C be nonempty sets and lef f map A to B and g map B to C.

If g of f maps A to C is an injection, then f is an injection.

Not sure how to do this proof

2. Originally Posted by dwsmith
Let A,B, and C be nonempty sets and lef f map A to B and g map B to C.

If g of f maps A to C is an injection, then f is an injection.

Not sure how to do this proof
How about an indirect proof, that is: show that if $f:\, A\rightarrow B$ is not an injection, then $g\circ f:\, A\rightarrow C$ is not an injection either.

3. Originally Posted by dwsmith
Let A,B, and C be nonempty sets and lef f map A to B and g map B to C.

If g of f maps A to C is an injection, then f is an injection.

Not sure how to do this proof
Here are some thoughts on an informal direct proof.
Let's say you already know (in the given context) the definition of injection.

For arbitrary x and y in A, suppose that over in B, f(x) = f(y).
This identity should tell you that (over in C) the image of f(x) under g is identical to the image of f(y) under g, i.e., that g(f(x)) = g(f(y)).
This can be written, (gof)(x) = (gof)(y).
Since gof is injective, you now have x = y (by definition).
So the arbitrariness of x, y and the supposition that f(x) = f(y), leads you to x = y, making f injective (by definition).