Error in my proof that $f \circ f^{-1} \subseteq I_B$

Question

I was marked off on this, and am thinking it has to do with the inverse relation not necessarily being a function?

Problem:

For non-empty sets $A$ and $B$, if $f: A\to B$ is a function, then $f \circ f^{-1} \subseteq I_B$

My Proof:

Let $f: A\to B$ and let $(a,b) \in f$. Then $(b,a) \in f^{-1}$. Then $f \circ f^{-1}(b)=f(f^{-1}(b))=f(a)=b$ giving $(b,b)\in f \circ f^{-1}\Rightarrow f \circ f^{-1}\subseteq I_B.$

Answer 1

Since you don't know that $f^{-1}$ is a function, you don't know that $f\circ f^{-1}$ is a function either. Therefore you can't use the notation $(f\circ f^{-1})(b)$ as if it necessarily means only a single thing.

Your argument therefore shows at best that if $(a,b)\in f$ then $(b,b)\in f\circ f^{-1}$, but not that all elements of $f\circ f^{-1}$ have this form.

A valid proof might start:

Let $(b_1,b_2)$ be an arbitary element of $f\circ f^{-1}$. By the definition of $\circ$ this means that there exists an $a$ such that $(b_1,a)\in f^{-1}$ and $(a,b_2)\in f$ ...