The cardinality of the preimage of an ordinal Define a cardinal as an ordinal $\kappa$ such that for all ordinals $\alpha < \kappa$, $\alpha$ injects into $\kappa$ but is not bijective to $\kappa$.
Let $\kappa$ be an infinite cardinal. I'm asked to show that there is a function $f:\kappa\to \kappa$ such that for all $\alpha < \kappa$, $|f^{-1}(\alpha)|=\kappa$.
Here's my idea: any infinite cardinal is in particular an infinite set. For any infinite set $\kappa$, there is a bijection $\phi$ from $\kappa$ to $\kappa\times \kappa$.  Consider the projection map $proj_1:\kappa\times\kappa\to\kappa, (x,y)\mapsto x$. Then its preimage is $\{x\}\times\kappa$, which is bijective to $\kappa$, and hence has cardinality $|\kappa|=\kappa$. This gives a map $\kappa\times\kappa\to\kappa$ such that for any $\alpha\in\kappa$, the preimage of $\alpha$ has size $\kappa$. If we compose this map with the $\phi$ mentioned above, would the preimage still have size $\kappa$? If so, how to prove this? And does the first part of the proof look okay? (One reason I'm not confident about it is that I don't see where I used the condition "for all ordinals $\alpha < \kappa$, $\alpha$ injects into $\kappa$".)
 A: 
I'm not confident about it is that I don't see where I used the condition "for all ordinals $\alpha<\kappa$, $\alpha$ injects into $\kappa$".

That $\alpha$ injects into $\kappa$ in your definition is superfluous: this is true for any $\alpha<\kappa$ (since $\alpha<\kappa$ implies $\alpha\subset\kappa$, so you could take the identity map). It is therefore not surprising that you don't need it in your proof.
More general, you do not necessarily need to use all parts of the definition of a cardinal, since you're proving that "if $\kappa$ is an infinite cardinal, then something something". There may be other objects for with something something is true as well, that aren't cardinals. You would only need to use all parts of the definition of a cardinal in claims of the form "if something something, then $\kappa$ is a cardinal".
In fact, it's true for any infinite set $A$ that there exists $f:A\to A$ such that $|f^{-1}(a)|=|A|$ for all $a\in A$. You do not need $A$ to be a cardinal.

If we compose this map with the $\phi$ mentioned above, would the preimage still have size $\kappa$?

Since $\phi$ is a bijection, it does not have any effect on the cardinality of the preimage. This is because bijections preserve cardinality by definition of cardinality: $A$ and $B$ have the same cardinality iff there exists a bijection $A\to B$. So the preimage $\{x\}\times\kappa$ of $x\in\kappa$ under $proj_1$ has the same cardinality as $\phi^{-1}[\{x\}\times\kappa]$, since $\phi^{-1}$ is a bijection.
