f is injective. Furthermore, the restriction of g on the image of f is injective. In particular, if the domain of g coincides with the image of f, then g is also injective. You can proof this by contradiction as Diana has already demonstrated.

You know that f must be 1-1. Suppose on the contrary it is not true that f is 1-1. Then there is an $x_1$ and $x_2$ such that $x_1 \ ne x_2$ but $f(x_1) =f(x_2)$ (think $f(x)=x^2$ on the real line). Since g is 1-1 g(f(x_1)) = g(f(x_2)) which implies x_1=x_2 oops violates f not being 1-1.

f is injective. Furthermore, the restriction of g on the image of f is injective. In particular, if the domain of g coincides with the image of f, then g is also injective. You can proof this by contradiction as Diana has already demonstrated.

Dear Diana, thank you for this proof. But what we can do in the case that f(x_1) is not in the domain of g?

Dear Ralf, I agree with you in the case that the domain of g coincides with the image of f. What we can say in general? (Please note that f(x_1) [and hence f(x_2)] does not belong to D_g necessarily.)

If say f(x_1) does not belong to D_g, then gof is not well-defined at all, since gof(x_1) =g(f(x_1)) is not defined. If the image of f is a proper subset of D_g, then you dot not have enough information to make a statement, i.e., g could be injective or not. Let A=im(f) denote the image f and B=D_g-im(f) the complementary set. If and only if g(A) and g(B) are disjunct AND the restriction of g on B is injective, then g is injective.

To proof the "if"-part of the statement, assume the x_1 and x_2 are in the domain of g.

If x_1 and x_2 are both in B, then f(x_1) \neq f(x_2), since the restriction of g on B is injective. If x_1 and x_2 are both in A, then f(x_1) \neq f(x_2) since the restriction of g on A is injective. If x_1 is in A and x_2 is in B(or vice versa) then f(x_1) \neq f(x_2) since g(A) and g(B) are disjunct.

To proof the "only if"-part, assume that g is injective. If g(A) an g(B) are not disjunct, then there exist y_1 in A and y_2 in B, such that g(y_1)=g(y_2) which is a contradiction to the injectivity. If the restriction of g on B is not injective, the g is obviously also not injective on D_g.

I naturally assumed that D was a domain that g(f) is well defined on. Apologies for not crossing my epsilons and dotting my deltas.

Let f(x) = x^3 and g(x) = sqrt(x). Then gof(x) = sqrt (x^3) is well-defined and say f(-1) does not belong to D_g.

I agree with you that g must be injective on the intersections of the sets im(f) and D_g (and this is true for any pair of arbitrary real functions f and g without restrictions on their domains or images).

Now, I think that we can make a statement like this: "GoF is injective if and only if F and G are injective on D_{GoF} and L, respectively." (L denotes the intersection of image(F) and domain(G) and we define D_{GoF} = The set of all x's in D_F such that F(x) belongs to D_g.)

Dear Mohammed, I agree with you. I would it formulate this way: GoF is injective if and only F is injective on D_{GoF} and G is injective on F(D_{GoF}).

Thank you Diana and Ralf. Now we have the excellent formulate of Ralf.

Diana, Instead of using g is 1-1 you should use gof is 1-1. Actually, g being a well-defined function, f(x_1) =f(x_2) implies g(f(x_1) = g(f(x_2)) and then the injectivity of gof implees x_1 = x_2 contradicting x_1 \neq x_2 . Very nice proof.