# Creation of limits and diagram chasing

Let T be a monad in X. I want to prove that the forgetful functor G from category of T-algebras on X to X creates limits.

I have read that it can be done via "diagram chasing". At this point, I am not sure that I know what exactly this means.

Suppose GH has a limit in X. To prove that G creates limits, I need to find a unique structure on LimGH and show that it will be a limit in the category of T-algebras.

How can I do this using diagram chasing? What is diagram chasing?

I am always stuck on problems which require diagram chasing, so I think I just don't understand the idea. Could you explain it somehow? Thanks!

• What is $T$? Is it a monad? Commented Apr 18, 2020 at 20:34
• @CaptainLama yes Commented Apr 18, 2020 at 20:36
• @jgon as in MacLane's CWM Commented Apr 18, 2020 at 21:02
• @arryn Nvm I managed to figure out what you meant by reading your question again.
– jgon
Commented Apr 18, 2020 at 21:03
• Just draw everything we know and seek for the arrows we need on the picture. For starters, assuming $H:D\to Alg_T(X)$, and let $L:=\lim GH$. There's an arrow $TL\to Td\to GHd$ for every $d\in Ob\, D$, and by the limit property they induce a $TL\to L$, that will give the $T$-algebra structure on $L$. Commented Apr 18, 2020 at 22:25

Note I'm not sure how to explain diagram chasing except by giving an example, so I'll go through the proof, and then try to explain how this is diagram chasing.

Setting up terminology:

Let $$X$$ be a category, and $$(T,\mu: T\circ T\to T,\eta : 1_X\to T)$$ a monad on $$X$$. Let $$H : I \to T\newcommand\Alg{\text{-}\mathbf{Alg}}\Alg$$ be a diagram of shape $$I$$. Let $$G:T\Alg \to X$$ be the forgetful functor. Let $$(L,\sigma)$$ be a limiting cone to $$GH$$ with $$L\in X$$ the vertex and $$\sigma_i : L\to GH_i$$ the morphisms of the cone.

We need to show that unique lift of this cone to $$T\Alg$$, and this lift will also be a limiting cone. Since the forgetful functor is faithful, this amounts to showing that there is a unique algebra structure on $$L$$ making $$(L,\sigma)$$ a cone in $$T\Alg$$ and then showing that this cone is a limiting cone.

The algebra structure on $$L$$

An algebra structure on $$L$$ is a morphism $$\alpha_L : TL\to L$$ such that $$\alpha_L\circ \eta_L = 1_L$$, and such that the square $$\require{AMScd} \begin{CD} T^2L @>T\alpha_L >> TL \\ @V\mu_L VV @V\alpha_L VV\\ TL @>\alpha_L >> L \end{CD}$$ commutes.

Now $$L$$ is the limit of $$GH$$, so morphisms $$\alpha_L$$ from $$TL$$ to $$L$$ correspond to a unique cone $$\sigma_i \circ \alpha_L : TL \to GH_i$$. (Conversely, a cone $$\pi_i : TL\to GH_i$$ gives such a morphism $$\alpha_L$$.)

Moreover, $$\alpha_L$$ is an algebra structure on $$L$$ making each $$\sigma_i$$ a morphism of $$T$$-algebras if and only if the following diagram commutes: $$\begin{CD} TL @>T\sigma_i >> TH_i \\ @V\alpha_L VV @V\alpha_{H_i}VV \\ L @>\sigma_i >> H_i. \end{CD}$$

Therefore in order for $$\sigma_i$$ to be a morphism of $$T$$-algebras, the $$T$$-algebra structure on $$L$$ must be given by the morphism $$\alpha_L : TL\to L$$ corresponding to the cone $$\alpha_{H_i}\circ T(\sigma_i)$$. In particular, the $$T$$-algebra structure on $$L$$ making $$(L,\sigma)$$ into a cone in $$T\Alg$$ is unique, if it exists.

To see that it exists, we need to show that $$\alpha_L$$ satisfies the properties required to be a monad algebra. For this, note that two morphisms to the limit $$L$$ are equal if and only if they are equal when you apply each $$\sigma_i$$.

So for the unit diagram, we need to show $$\alpha_L \eta_L =1_L$$. Applying $$\sigma_i$$, we get $$\sigma_i\alpha_L\eta_L = \alpha_{H_i}(T\sigma_i)\eta_L = \alpha_{H_i}\eta_{H_i}\sigma_i = \sigma_i=\sigma_i 1_L,$$ with the first equality the definition of $$\alpha_L$$, the second equality is naturality of $$\eta$$, and the third is the unit identity for $$H_i$$ as a $$T$$-algebra.

For the multiplication compatibility square, we need to show $$\alpha_L(T\alpha_L) = \alpha_L\mu_L$$. Again, we apply $$\sigma_i$$. We get \begin{align} \sigma_i\alpha_L(T\alpha_L) &= \alpha_{H_i}(T\sigma_i)(T\alpha_L) \\ &= \alpha_{H_i}(T(\sigma_i\alpha_L)) \\ &= \alpha_{H_i}(T(\alpha_{H_i}(T\sigma_i))) \\ &= \alpha_{H_i}(T\alpha_{H_i})(T^2\sigma_i) \\ &= \alpha_{H_i}\mu_{H_i}(T^2\sigma_i) \\ &= \alpha_{H_i}(T\sigma_i)\mu_L \\ &= \sigma_i\alpha_L\mu_L, \end{align} as required. I'm not going to add anything further explaining these steps in detail. Instead, I'll say that these come from a geometric observation about a cubical diagram we can draw. The observation is that we know all the faces except the $$L$$ face commute, either by naturality, construction of $$\alpha_L$$, or the fact that $$H_i$$ is a $$T$$-algebra, so we can push the paths we care about around the cube across the faces that commute, and thereby demonstrate that they are equal. Diagram drawn with tikzcd.

Thus $$\alpha_L$$ does define a $$T$$-algebra.

An aside on diagram chasing

Diagram chasing is a vaguely defined term. It refers to a style of argument, like the one just given, where to prove things, you draw a diagram of objects and morphisms, and then either move elements around the diagram, construct new morphisms, or use commutativity of the diagram (where it is commutative) to prove what you want to prove. I'm not sure there's a more specific definition I could give.

Back to the problem

We have now shown that there is a unique lift of a limiting cone to $$T\Alg$$ by producing a unique algebra structure on $$L$$. All that's left is to show that this lifted cone is also a limiting cone.

Since it's a limiting cone in $$X$$, this comes down to just showing that if $$(A,\pi)$$ is a cone to $$H$$ in $$T\Alg$$, then the induced map $$p: A\to L$$ in $$X$$ is actually a map of $$T$$-algebras. In other words, you need to show that $$\begin{CD} TA @>Tp >> TL \\ @V\alpha_A VV @V\alpha_L VV\\ A @>p>> L \\ \end{CD}$$ commutes. The solution is very similar to what we did above, so I'll leave this as an exercise.