Why are the morphisms of the category of sets functions? Shouldn't the morphisms take an object in a category and turn it into another object of the category, i.e. map Set to Set. I don't understand how e.g. $f(x)=x^2$ is a map from a set to a set.
If the morphism had been the image of a set under a function it would make more sense to me.
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Shouldn't the morphisms take an object in a category and turn it into another object of the category
No, what makes you think so? Are you perhaps confusing morphisms with functors?
Morphism is just an arrow between two objects. Objects in set category are just sets. If $A,B$ are two sets then what would be an arrow $A\to B$? The natural choice is a function.
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I don't understand how e.g. $f(x)=x^2$ is a map from a set to a set.
In fact it is a map from a set to a set: you haven't given a precise definition of the function, so there are multiple options what the sets are, but $f$ may, e.g., map from the domain $\mathbb R$ (the set of all real numbers) to the codomain $\mathbb R_{\geq 0}$ (the set of all nonnegative real numbers). Then, in the category of sets, $f$ would constitute a morphism between the objects $\mathbb R$ and $\mathbb R_{\geq 0}$.
In this context, you probably want to be thinking of functions as rules taking an input value from one set and returning an output value from another set.
This is what is called a functor.
A morphism is (more or less) exactly a generalization of functions between sets. One function corresponds to one morphism / arrow.
Your example $f(x) = x^2$ might be considered as a morphism $f: \mathbb{R} \rightarrow \mathbb{R}$. Sometimes we also write $f \in Hom(\mathbb{R}, \mathbb{R})$ or $f \in Mor(\mathbb{R}, \mathbb{R})$, but those are just different notations for the same thing.
Morphisms are one of the building blocks of categories. Each category $C$ as some objects $O \in Ob(C)$, and for each pair of objects $O, P \in Ob(C)$, there is a set of morphisms $Mor(O, P)$.
Does that make something more clear?