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Let $f: \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$ be defined by $f((a,b))=b$.

I am confused as to whether $f \circ f$ exists here.

I was taught that for this composite function to exist, the codomain of the inner function must be equal to the domain of the outer function. Now I understand $\mathbb{Z}$ is not equal to $\mathbb{Z} \times \mathbb{Z}$, however since $f$ in this case doesn't even use $a$, is it at all possible for $f \circ f$ to indeed exist?

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    $\begingroup$ No, it is not defined, because of the reason you already mentioned. $\endgroup$
    – geetha290krm
    Commented Apr 16 at 0:18
  • $\begingroup$ Got it - just wanted to confirm, thanks! I think I was just overthinking it $\endgroup$
    – scob_
    Commented Apr 16 at 0:25
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    $\begingroup$ Can't exist. For the reason you gave. Not using something doesn't mean you don't have to put something in. $\endgroup$
    – fleablood
    Commented Apr 16 at 0:26
  • $\begingroup$ Now it's possible to have a function $g:\mathbb Z \to \mathbb Z\times Z$ that is defined as $g(b) = (7, b)$. Then we can have $f\circ g\circ f$. And what's more $f\circ g \circ f = f$. If we input $(a,b)$ then $f:(a,b)\to b$. And then $g: b \to (7,b)$ and then the second $f:(7,b)\to b$. So $f\circ g\circ b: (a,b)\to b$ and is, thus, the exact same function as $f$..... but that wasn't your question, was it? $\endgroup$
    – fleablood
    Commented Apr 16 at 0:32

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No.

Given two functions $f: A \to B$ and $g: C \to D$, we can only form $g \circ f: A \to D$ if $B \subseteq C$.

In your example, $\mathbb{Z}$ is not a subset of $\mathbb{Z} \times \mathbb{Z}$, so this composition is not defined. However, if you come up with some $h: \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$, like say by fixing some $m \in \mathbb{Z}$ and defining the inclusion $\iota_m(n) = (m, n)$, then you can form the composition $f \circ \iota_m \circ f$ $$ \mathbb{Z} \times \mathbb{Z} \xrightarrow{f} \mathbb{Z} \xrightarrow{\iota_m} \mathbb{Z} \times \mathbb{Z} \xrightarrow{f} \mathbb{Z} $$ that maps $(a, b) \mapsto b \mapsto (m, b) \mapsto b$ for $(a, b) \in \mathbb{Z} \times \mathbb{Z}$, so it turns out that $f \circ \iota_m \circ f = f$.

However, that is certainly not the only possible map $\mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ (even if we only consider abelian group homomorphisms).

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    $\begingroup$ I think it may be important to note that $\iota_m$ will always append the same $m$ no matter what. If $f:(a,b)\to b$ and $a\ne m$ the $a$ is simply lost for ever. And there is no function $MAGIC:\mathbb Z \to \mathbb Z\times Z$ that we "rembember" that the $b$ was once attached to $a$ and map $MACIG:b\to (a,b)$. All $\iota_m$ does is map any $x$ to $(m,x)$ no matter what.... in other words there is no "inverse" $f^{-1}$ so that $f^{-1}\circ f$ will map $(a,b)\to b \to (a,b)$. That just is not possible.... Or maybe it's not important to note that. $\endgroup$
    – fleablood
    Commented Apr 16 at 0:40

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