Given that f : A(arrow)B and g : B(arrow)C are functions. In each of the followi

Question

Given that f : A(arrow)B and g : B(arrow)C are functions. In each of the following cases, answer yes or no. If "yes", give a proof. If the answer is "no", give a counterexample and say what additional hypotheses are needed to make the statement true; then prove the statement with the additional hypotheses. (c represents the composition symbol)

1. If g (c) f is one-to-one, must f be one-to-one?

2. If g (c) f is one-to-one, must g be one-to-one?

3. If g (c) f is onto, must f be onto?

4. If g (c) f is onto, must g be onto?

Explanation / Answer

1.

True,

gof is 1-1.

Suppose x,y be two elements in A such that f(x) = f(y).

Then gof(x) = g[f(x)] = g[f(y)] (since f(x) = f(y) ) = gof(y)

=> x = y, since gof is 1-1.

This shows that f(x) = f(y) => x=y, i.e f is 1-1.

2.

False, as an example take

A = {1,2}, B = {1,2,3}, C = {1,2,3}

f(1) = 1, f(2) = 2, g(1) = g(3) = 1, g(2) = 2.

Then gof(1) = 1, gof(2) = 2, so gof is 1-1.

But g is not 1-1.

If in addition to gof being 1-1 we have f is onto, then g is 1-1.

Let x,y be two elements in B such that g(x) = g(y)

Then since f is onto there exists x', y' such that f(x') = x, f(y') = y.

Now gof(x') = g[f(x')] = g(x) = g(y) = g[f(y')] = gof(y')

since gof is 1-1, this implies x'=y' => x = y.

Thus g(x) = g(y) => x = y, i.e g is 1-1.

3.

False, as an example take A = {1,2}, B = {1,2,3}, C = {1,2}

f(1) = 1, f(2) = 2, g(1) = 1, g(2) = 2, g(3) = 1.

gof(1) = 1, gof(2) = 2.

So gof is onto, whereas f is not onto.

If in addition to gof being onto we have g is 1-1, then f is onto.

Let b be an element of B. Suppose if possible b is not in the range of f.

Also let g(b) = c, then for any element x in B distinct from b, g(x) is distinct from c.

Now we claim gof does not have c in its range and hence a contradiction to the assunmption that gof is onto.

Suppose if possible gof(a) = c, then g[f(a)] = c = g(b) => f(a) = b, in other words b is in the range of f, a contradiction to our assumption that b is not in the range.

This shows that our assumption is false, in other words f is onto.

4.

True,

gof is onto.

Let c be an element of C.Then since gof is onto there exists a in A such that gof(a) = c.

Now if we call f(a) as b, then g(b) = gof(a) = c.

Thus g is onto.

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