I am learning basics from Category theory and in Steve Awodey's book "Category Theory" I've found the following:
If you think of a functor $F : C \to D$ as “picture” of $C$ in $D$, then you can think of a natural transformation $\varphi_C : FC → GC$ as a “cylinder” with such a picture at each end.
Can someone help me understand where did this "cylinder" come from? How to visualize it?
As always, to get an understanding of abstract definitions one should consider examples. The easier, the better.
Take $C=\bullet$ the category with one object and no nontrivial morphisms and $D=\mathsf{Set}$. A functor $F:C\rightarrow D$ does nothing more than picking a set $F(\bullet)$. A natural transformation $\alpha:F\Rightarrow G$ just consists of a function $\alpha:F(\bullet)\rightarrow G(\bullet)$. In some sense a line is a cylinder, isn’t it?
If you are not convinced, take the next best example, where $C=(0\rightarrow 1)$ the free walking arrow (no nontrivial morphisms besides the depicted one). A functor picks two sets $F(0)$ and $F(1)$ together with a function $f:F(0)\rightarrow F(1)$. A natural transformation amounts to two functions $\alpha_0:F(0)\rightarrow G(0)$ and $\alpha_1:F(1)\rightarrow G(1)$ making the square $$\begin{array}{ccc} F(0)&\overset{\alpha_0}\longrightarrow & G(0)\\ \downarrow&&\downarrow\\ F(1)&\overset{\alpha_1}\longrightarrow & G(1) \end{array}$$ commute. In some sense, a square can be regarded as a cylinder, doesn’t it?
If you are still not convinced, take the next best example, where $$C=\begin{array}{ccc} 0&\rightarrow&1\\&\searrow&\downarrow\\&&2\end{array}$$ I leave it as an exercise to spell out what a functor and a natural transformation look like. Can you see how this is a cylinder?
Finally, as some kind of foreshadowing, let me explain, how a natural transformation can be formally considered as a cylinder: A natural transformation can be described as a functor $(0\rightarrow 1)\rightarrow\operatorname{Fun}(C,D)$. Using cartesian closedness of $\mathsf{Cat}$ this is the same as a functor $(0\rightarrow 1)\times C \longrightarrow D$. If you think of $(0\rightarrow 1)$ as a line, $(0\rightarrow 1)\times C$ is a cylinder with base-shape $C$. The functor takes that to its image inside $D$.