Is $g \circ f$ invertible if both $f$ and $g$ are invertible?.
I know that $f: A \rightarrow B$ and $g: B \rightarrow C$.
This is what I have so far:
If $f$ is invertible, $f^{-1} : B \rightarrow A$, $f^{-1} \circ f = id_A$. And if $g$ is invertible, $g^{-1} : C \rightarrow B$, $g^{-1} \circ g = id_B$.
Given all of this information, however, I am not sure how this proves that $g \circ f$ is invertible.
What you need to use is the fact that co.position of functions is associative, and then $(g \circ f)^{-1}= f^{-1} \circ g^{-1}$ . We can check this as follows:
$(f^{-1} \circ g^{-1}) \circ (g\circ f)= f^{-1} \circ (g^{-1} \circ g)\circ f= (f^{-1})\circ id \circ (f)= f^{-1} \circ f= id$