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I cannot follow the definition of Hom Functors. What is the meaning when we write

$${\rm Hom}(X,-) : C \to {\rm Set}$$

or

$$f \circ - : {\rm Hom}(X,A) \to {\rm Hom}(X,B)$$

And how can we define locally small categories?

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    $\begingroup$ Have you looked at any introduction to category theory? $\endgroup$ – Martin Brandenburg Dec 11 '13 at 15:01
  • $\begingroup$ Yes, but almost confusing statements, I am new to this subject and having many difficulties to have an understanding of this course. $\endgroup$ – S786 Dec 11 '13 at 15:02
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In a category $C$ let us fix an object $X$. For any object $Y$ in $C$ there is a set (more generally a class, but if $C$ is a locally small category, then it is a set) of all morphisms from $X$ to $Y$, called hom-set Hom$(X,Y)$, though it also appears as $C(X,Y)$ or Hom$_C(X,Y)$. Of course there need not be an arrow from $X$ to $Y$, and in that case the set is just the empty set.

So this assigns to every object $Y$ in $C$ the set Hom$(X,Y)$. In order to have a functor we need an arrow function sending arrows in $C$ to arrows in Set in such a way that an arrow $f:A\to B$ is sent to a set map Hom$(X,f):$Hom$(X,A)\to$Hom$(X,B)$. This means that we want to assign to each arrow $g:X\to A$ and arrow $X\to B$. There is a natural way to do this, namely by composing this $g$ with the arrow $f$. So we get $$\text{Hom}(X,f):\text{Hom}(X,A)\to\text{Hom}(X,B)\\g\mapsto f\circ g$$ This defines the "arrow function" of the Hom functor.
Now there are two conditions this function must satisfy to be a functor. One of them is that for arrows $f:A\to B$ and $h:B\to C$ we have $\text{Hom}(X,hf)=\text{Hom}(X,h)\circ\text{Hom}(X,f)$. In other words, given a $g:X\to A$, composing this with $hf$ yields the same arrow $X\to C$ as if we first composed with $f$, then composed $fg$ with $h$. Can you formally check that this gives the same arrow? Compare with the axioms for a category!

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