I did it like this:Suppose that f:A $\displaystyle \rightarrow$ B and g:B$\displaystyle \rightarrow$C are mappings. Show that:

1. f and g are both injective $\displaystyle \Rightarrow$ g $\displaystyle \circ$f is injective

Let $\displaystyle a \in A$, $\displaystyle b \in B$ and $\displaystyle c \in C$.

If f and g are injective $\displaystyle \forall$ x,y $\displaystyle \in$A, f(x)=f(y) $\displaystyle \Rightarrow$ x=y.

Therefore: f(a)=b and g(b)=c where a is unique to b and b is unique to c (one to one function by the injective property).

$\displaystyle g \circ f(x)=g(f(x))$

Let x=a:

$\displaystyle g \circ f(a)=g(f(a))=g(b)=c$

Since a is unique to be and b is unique to c then $\displaystyle g\circ f(a)$ is also a one to one function. Hence $\displaystyle g \circ f$ is injective.

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My issue here is the part where I mention that f(a)=b and g(b)=c. Is this a valid claim?