I am reading "Category Theory" (2nd ed.) of Awodey, and I'm stuck at page 96 (proposition 5.12) when pullbacks are presented as functors:

The pullback under question corresponds to this square:

$C' \times_C A \rightarrow_{h'} \hspace{1ex}A$

$\hspace{1em}\downarrow_{\alpha'} \hspace{6ex} \downarrow_\alpha$

$\hspace{1em}C' \hspace{3ex} \rightarrow_h \hspace{1ex} C$

Here is the statement of Awodey's book that I do not understand:

> Pullback is a functor. That is, for fixed $C' \rightarrow_h C$ in a category $\mathbf{C}$ with pullbacks, there is a functor
> 
> $h^* : \mathbf{C}/C \rightarrow \mathbf{C}/C'$
>
> defined by
>
> $(A\rightarrow_\alpha C) \mapsto (C'\times_C A \rightarrow_{\alpha'} C')$
>
> where $\alpha'$ is the pullback of $\alpha$ along h

The problem that I see is that, given initially:

$\hspace{12ex}A$

$\hspace{12ex} \downarrow_\alpha$

$\hspace{1em}C' \hspace{3ex} \rightarrow_h \hspace{1ex} C$

there can be several pullbacks on it, for example, in addition to $(\alpha',h')$, there could be $(\alpha_2',h_2')$, and the unique condition is that there exist an isomorphism i such that $\alpha_2' = \alpha\circ i$ and $h_2' = h'\circ i$.
Worse, given a pullback $(\alpha',h')$, one can build as much as pullbacks as there exist isomorphisms, as given any isomorphism j (with domain $C' \times_C A \rightarrow_{h'}$) the two arrows $(\alpha'\circ j,h' \circ j)$ form a new pullback.

So, how could we build a functor if the image arrow is only defined up to an arbitrary isomorphism?