Why is the composition of a surjective and injective function neither surjective nor injective?

Question

I am currently preparing for an exam coming up and I was therefore looking in earlier exam-sets, and found a question with a solution I just cannot make sense in my head.

Let $A, B, C$ be three arbitrary sets, $f\colon B\to C$ be a surjective function and $g\colon A \to B$ be an injective function. Then the composition $f\circ g$ is?

1. Injective
2. Surjective
3. Bijectie
4. Neither injective nor surjective

The answer-sheet suggests that the composition of these two functions should neither be injective nor surjective. Whenever I try drawing functions by example however, it seems easy to make a composition that is both injective and surjective.

Thanks.

Answer 1

Of course you can construct examples where the composition is injective, or where it is surjective. But the point is that you can also construct examples where neither of these properties is fulfilled. The question has to be interpreted in the way "Can one conclude from $f$ being surjective and $g$ being injective that the composition is necessarily surjective, etc.?'' And this is not the case.

Answer 2

You can make the composition both injective and surjective (the easiest example is to take $f$ and $g$ to both be the identity map on $A = B = C$). The point is that you can't conclude, from the data provided, that it's either. The easy way to see this is to have one of $f$ and $g$ be the identity, and the other either surjective but not injective or injective but not surjective: then the composition is just the same as the non-identity one, which doesn't have the relevant property.

Answer 3

The answer to the question, as it is written, should probably be "There is not enough information to conclude anything about the injectivity or surjectivity of the composition." However, as this is clearly not the intended answer (as it is not an option in your list of multiple choices), we have to get into the heads of the question writers and try to figure out what they meant to ask. My reading of the question is that it would better be phrased as

Let $A$, $B$, $C$ be three arbitrary sets, $f : B \to C$ be a surjective function and $g : A \to B$ be an injective function. Which of the following statements is true:

1. $g\circ f$ is always injective.
2. $g\circ f$ is always surjective.
3. $g\circ f$ is always bijective.

In this phrasing, the correct answer is that all three statements are false. Again, there isn't enough information to conclude anything about the surjectiveity or injectivity of the composition. To see this, we need only construct two counterexamples:

• Let $A = \{\ast\}$ be a singleton set, and let $B = C = \{ a, b \}$ be a set with two elements. Define $g : A \to B$ by $g(\ast) = a$, and take $f : B \to C$ to be the identity map. Then the composition $f\circ g$ is injective (any map from a singleton set to any other set must be injective), but is not surjective ($b$ is not the image of any point in $A$ under the composition). Therefore the composition needn't be surjective (which also means that it needn't be bijective, as every bijective map is necessarily surjective).

• Let $A = B = \{a, b\}$ be a set with two elements, and $C = \{\ast\}$ be a singleton set. Let $f : A\to B$ be the identity map, and let $g : B\to C$ be the only possible map, i.e. the function defined by $g(x) = \ast$ for all $x\in B$. Then the composition $f\circ g$ is surjective, but not injective. Therefore the composition needn't be injective.

Note that we can construct examples of sets $A$, $B$, and $C$, and maps $f : A\to B$ and $g : B \to C$ such that the composition $f\circ g$ is injective (but not surjective), surjective (but not injective), or bijective. However, simply knowing that $f$ is injective and that $g$ is surjective is not enough to tell us anything about the properties of the composition.