I think I understand what a commutative square is and also what a natural transformation means.

Yet I cannot think of an example of a transformation that is not natural. I’ve found some papers online that talk about some transformations that are not natural, but they don’t serve well as examples since they are way too advanced for me.

Is there some simple or even canonical example of transformation that generates a non-commutative diagram?

  • $\begingroup$ What do you mean by a "transfornation"? The noun phrase "natural transformation" can't be decomposed as an application of the adjective "natural" to the noun "transformation" in the definitions of these terms that I know. $\endgroup$
    – Rob Arthan
    May 8, 2018 at 21:51
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    $\begingroup$ Transformation, for each object of the category create its component, all the components are the transformation. $\endgroup$
    – gurghet
    May 8, 2018 at 21:54

2 Answers 2


Let us construct a counterexample which in a certain sense is a "universal" example of the possible failure to be natural.

Here, $\mathcal{B}$ will be the category with two objects which we will call $X, Y$. Its morphisms are $\renewcommand{id}{\operatorname{id}}\id_X : X \to X$, $\id_Y : Y \to Y$, and another morphism we will call $f : X \to Y$; and the composition operation is constructed in the obvious way.

Similarly, $\mathcal{C}$ will be the category with four objects which we will formally call $F(X)$, $F(Y)$, $G(X)$, and $G(Y)$. Its morphisms will be $\id_{F(X)} : F(X) \to F(X)$, $\id_{F(Y)} : F(Y) \to F(Y)$, $\id_{G(X)} : G(X) \to G(X)$, $\id_{G(Y)} : G(Y) \to G(Y)$; then morphisms which we will formally call $F(f) : F(X) \to F(Y)$; $\alpha_Y : F(Y) \to G(Y)$; $G(f) : G(X) \to G(Y)$; $\alpha_X : F(X) \to G(X)$; $\alpha_Y \circ F(f) : F(X) \to G(Y)$; and $G(f) \circ \alpha_X : F(X) \to G(Y)$. Again, the composition will be constructed in the obvious way.

Now, we can construct functors $F : \mathcal{B} \to \mathcal{C}$ and $G : \mathcal{B} \to \mathcal{C}$ as suggested above: $F$ on objects sends $X$ to the object we formally called $F(X)$, and $Y$ to $F(Y)$, and on morphisms sends $\id_X$ to $\id_{F(X)}$, $\id_Y$ to $\id_{F(Y)}$, and $f$ to $F(f)$. $G$ is constructed similarly. We also construct $\alpha : F \to G$ which sends $X$ to $\alpha_X$ and $Y$ to $\alpha_Y$. This fails to be a natural transformation since in our category $\mathcal{C}$, $G(f) \circ \alpha_X$ and $\alpha_Y \circ F(f)$ are constructed to be two different morphisms.

  • $\begingroup$ Thanks Daniel, this is a very good canonical example. $\endgroup$
    – gurghet
    May 8, 2018 at 22:17
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    $\begingroup$ This example is actually quite profound it took me a while to really get it $\endgroup$
    – gurghet
    May 9, 2018 at 18:57
  • $\begingroup$ It is a great example! It is interesting that we don't have a good name for $\alpha$ sending objects in one category to morphisms in another category. $\endgroup$ Jan 29, 2019 at 6:37

It's REALLY EASY to get a non-natural transformation (much easier than it is to get a natural transformation!). Just choose a morphism for each object completely randomly, and the transformation you get will almost certainly be non-natural. Seriously, try it: take your favorite pair of functors $F,G:C\to D$, and for each object $c$ of $C$, pick some map $\alpha_c:Fc\to Gc$ in some completely arbitrary way (not in some consistent way for all $c$). The result will pretty much never be a natural transformation.

For a simple explicit example using familiar categories, you could take your favorite natural transformation and just change one of the maps in some arbitrary way. For instance, let's start with the identity natural transformation from the identity functor $Set\to Set$ to itself. That's just the natural transformation which to every set $X$ associates the identity map $X\to X$. Naturality is just the statement that for any map $f:X\to Y$, the diagram $$\require{AMScd} \begin{CD} X @>{Id_X}>> X\\ @V{f}VV @V{f}VV \\ Y @>{Id_Y}>> Y \end{CD}$$ commutes.

Now let's modify this natural transformation, so that it gives the identity map for every set except $\mathbb{N}$, for which it gives the constant map $c$ with value $5$. This transformation is no longer natural. For instance, if $f:\mathbb{N}\to\mathbb{N}$ is the map $x\mapsto x+1$, then the diagram $$\require{AMScd} \begin{CD} \mathbb{N} @>{c}>> \mathbb{N}\\ @V{f}VV @V{f}VV \\ \mathbb{N} @>{c}>> \mathbb{N} \end{CD}$$ does not commute ($f\circ c$ sends every number to $6$ and $c\circ f$ sends every number to $5$).

  • $\begingroup$ Set is a bit boring, but yes, this clarifies a lot. $\endgroup$
    – gurghet
    May 8, 2018 at 22:04

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