# True/false? $A,B,C$ are arbitrary sets and $f: A \rightarrow B$ and $g: B \rightarrow C$ are injective mappings. Then also $g \circ f$ is injective.

True/false? $A,B,C$ are arbitrary sets and $f: A \rightarrow B$ and $g: B \rightarrow C$ are injective mappings. Then also $g \circ f$ is injective.

There isn't a proof required, just a true or false and a quick counter example in case it's false.

I'm not sure but I say it's false.

Let $f(x) = 0$ and $g(x) = x$, so then $g$ is injective but in composition with $f$, so $g \circ f$, it won't be injective.

What do you think about that, is it correct?

• But f(x) = 0 is not injective. Commented Mar 16, 2017 at 8:12
• "Injective" means "two separate things in the domain will get taken to two separate things in the range". Do that once for $f$ and once for $g$; is the result necessarily still two separate things in the range of $g \circ f$? Commented Mar 16, 2017 at 8:21

## 2 Answers

The statement is true.

It's also very very easy to prove that it's true, and I encourage you to do it. Remember, $f\circ g$ is injective if, for any $a_1,a_2\in A$ such that $(f\circ g)(a_1)=(f\circ g)(a_2)$, you can prove that $a_1=a_2$.

In your proof, all you need are the following things:

• $(f\circ g)(a)=f(g(a))$ by definition
• $g$ is injective, meaning that if $g(b_1)=g(b_2)$, then $b_1=b_2$ for any $b_1,b_2\in B$
• $f$ is injective, meaning that if $f(a_1)=f(a_2)$, then $a_1=a_2$ for any $a_1,a_2\in A$
• $f(a)$ is an element of $B$ for every $a\in A$.

In your "counterexample", $f$ is not injective, so you did not disprove the original statement. Or, technically, $f$ can be injective if $A=\{0\}$, but in that case, $f\circ g$ is also injective.

$(g\circ f)$ is injective.

Let : $$A = \lbrace a_i \rbrace$$

$$B = \lbrace b_j \rbrace$$

$$C = \lbrace c_k \rbrace$$

$(g\circ f): (f: A \rightarrow B) \rightarrow C$

Since $f: A \rightarrow B$ is injective, i.e. $a_i \implies b_j ~~~\text{(Exactly one)}$ .

Since $g: B \rightarrow C$ is injective, i.e. $b_j \implies c_k ~~~\text{(Exactly one)}$ .

For $(g\circ f)$, $(a_i \implies b_j ~~~\text{(Exactly one)})\implies c_k$

Therefore $(g\circ f)$ is also one-one mapping.