Find limits and co-limits of diagrams over Vector

category-theorylimits-colimitslinear algebraquotient-spaces

I am having trouble understanding how to find limits and colimits of specific diagrams over the category of finite dimension vector spaces. I understand the definitions of cones, terminal objects, limits and colimits of diagrams as well as what it means for a diagram to commute. However I am unable to solve the following problem.

I need to find the limit and colimit of the following diagram:

$\require{AMScd}$
\begin{CD}
\mathbb{R}^2 \\
@AfAA\\
\mathbb{R}^2 @>g>> \mathbb{R}^2\\
\circlearrowright h
\end{CD}

where $f = \begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}$, $g = \begin{bmatrix}2 & 0\\0 & 0\end{bmatrix}$ and $h = \begin{bmatrix}0 & 0\\0 & 2\end{bmatrix}$

I understand that to find the limit, I need to find a vector space $V$ along with morphisms $v_1$, $v_2$ and $v_3$ such that the diagram commutes and that such a cone is the terminal object in the category of all commuting cones.

\begin{CD}
\mathbb{R}^2 @<v_1<< V\\
@AfAA \swarrow v_2 @VVv_3V\\
\mathbb{R}^2 @>g>> \mathbb{R}^2\\
\circlearrowright h
\end{CD}

I know that for commutativity we need $v_2 = h \circ v_2$, $v_1 = f \circ v_2$ and $v_3 = g \circ v_2$.

The colimit would be:

\begin{CD}
\mathbb{R}^2 @>w_1>> W\\
@AfAA \nearrow w_2 @AAw_3A\\
\mathbb{R}^2 @>g>> \mathbb{R}^2\\
\circlearrowright h
\end{CD}

My professor has given hints that to find the limit I should start with the product of all objects in the diagram and create a quotient space $V= \frac{\mathbb{R}^2 \times \mathbb{R}^2 \times \mathbb{R}^2}{?}$ as the apex of the cone. Similarly, for the colimit should be a quotient space of the direct sum $W= \frac{\mathbb{R}^2 \oplus \mathbb{R}^2 \oplus \mathbb{R}^2}{?}$. However, I do not know how to proceed (I know what quotient spaces are, but am not sure how to construct a specific one here). Furthermore, I should also prove, that the cone i find is actually the limit (colimit) of the diagram, which I also wouldn't know how to do.

I would be grateful for a thorough step-by-step explanation of the process of finding the limits are colimits of this diagram (and by extension diagrams of this type). The textbooks I have only give a very abstract overview of the subject and I suspect my inability to solve this is due to a poor understanding of constructing vector spaces with desired properties.

Best Answer

$\require{AMScd}\newcommand\mat[1]{\begin{bmatrix}#1\end{bmatrix}}$Let \begin{CD} V @= V @= V\\ @V v_1 VV @Vv_2VV @VV v_3 V\\ \Bbb R^2@<<f<\Bbb R^2@>>g>\Bbb R^2 \end{CD} be a cone. Then $h\circ v_2=v_2$, hence $(h-1)\circ v_2=0$. Since $h-1=\bigl[\begin{smallmatrix}-1&0\\0&1\end{smallmatrix}\bigr]$, we have $\det(h-1)\neq 0$, henec $v_2=0$. Consequently, $v_1=f\circ v_2=0$ and $v_3=g\circ v_2=0$. Hence there exists one and only one morphism making the following diagram commutative \begin{CD} V @= V @= V\\ @VVV@VVV @VVV\\ \{0\} @= \{0\} @= \{0\}\\ @VVV @VVV @VVV\\ \Bbb R^2@<<f<\Bbb R^2@>>g>\Bbb R^2 \end{CD} and this proves \begin{CD} \{0\} @= \{0\} @= \{0\}\\ @VVV @VVV @VVV\\ \Bbb R^2@<<f<\Bbb R^2@>>g>\Bbb R^2 \end{CD} to be a limit cone.

On the other hand, let \begin{CD} \Bbb R^2@<f<<\Bbb R^2@>g>>\Bbb R^2\\ @Vw_1VV@Vw_2VV@VVw_3V\\ W@=W@=W \end{CD} be a cocone. Then $w_1\circ f=w_3\circ g$, hence \begin{align} 0 &=w_1\circ f-w_3\circ g\\ &=\mat{w_1&w_3}\mat{f\\-g}\\ &=\mat{u_1&v_1&u_3&v_3}\mat{1&0\\1&1\\-2&0\\0&0}\\ &=\mat{u_1+v_1-2u_3&v_1} \end{align} from which $v_1=0$ and $u_1=2u_3$, so that \begin{align} \mat{w_1&w_3} &=\mat{u_1&v_1&u_3&v_3}\\ &=\mat{2u_3&0&u_3&v_3}\\ &=\mat{u_3&v_3} \mat{2&0&1&0\\0&0&0&1} \end{align} from which \begin{align} w_1&=\mat{u_3&v_3}\mat{2&0\\0&0}=w_3\circ g& &w_3=\mat{u_3&v_3}\mat{1&0\\0&1}=\mat{u_3&v_3} \end{align} Moreover, $w_2\circ h=w_2$ gives $w_2\circ(h-1)=0$ from which \begin{align} 0 &=w_2\\ &=w_3\circ g\\ &=\mat{u_3&v_3}\mat{2&0\\0&0}\\ &=\mat{2u_3&0} \end{align} which gives $u_3=0$, hence $w_3=\mat{0&v_3}=v_3\circ\mat{0&1}$. Since $\mat{0&1}$ is an epimorphism, $v_3$ is the only morphism making the following diagram commutative \begin{CD} \Bbb R^2@<f<<\Bbb R^2@>g>>\Bbb R^2\\ @V0VV@V0VV@VV\mat{0&1}V\\ \Bbb R@=\Bbb R@=\Bbb R\\ @Vv_3VV@Vv_3VV@VVv_3V\\ W@=W@=W \end{CD} and this proves \begin{CD} \Bbb R^2@<f<<\Bbb R^2@>g>>\Bbb R^2\\ @V0VV@V0VV@VV\mat{0&1}V\\ \Bbb R@=\Bbb R@=\Bbb R \end{CD} to be a colimit cocone.

Related Solutions

Trying to understand the definition of a universal property

First of all, that wikipedia definition is not great. You should however go back to it once your wrap your head around all this stuff.

When one first encounters the definition of a universal construction, it's a bit weird. The true way to understand this concept is with examples: just keep looking at examples until it makes sense.

Let's start with a simple example. Let $X, Y$ be sets. Then (one possible way) I can define the disjoint union is $$ X \amalg Y = \{(x, 0), (y, 1) \mid x \in X, y \in Y \} $$ and we can define injection morphisms $$i_X: X \to X \amalg Y \qquad i_X(x) = (x, 0)\\ i_Y: Y \to X \amalg Y \qquad i_Y(y) = (y, 1) $$ Now suppose I have a function $f: X \to Z$ and $g: Y \to Z$. Then I can construct a unique (this is really key here) map $h: X \amalg Y \to Z$ where $$ h(x, 0) = f(x) \text{ and } h(y, 1) = g(y) $$ Why is it unique? Because the definition depends directly on $f$ and $g$. So what this really gives me is this diagram

Given the arrows $f:X \to Z, g: Y \to Z$, we can get a unique arrow $h: X \amalg Y \to Z$. The forced existence of $h$ is indicated by the dashed arrow.

This is an example of a universal construction. Actually, this is an example of a colimit, which I will explain. This happens in so many places in math that there's a name for it, and thats what this universal arrow stuff is about.

So suppose $\mathcal{C}$ and $\mathcal{D}$ are categories with a functor $F: \mathcal{C} \to \mathcal{D}$. A universal morphism from an object $D$ to the functor $F$ is a pair $(C, u: D \to F(C))$ such that, for any $f: D \to F(C')$, the following diagram holds.

So if you have a morphism $f: D \to F(C')$, you automatically get a morphism $h: C \to C'$.

A colimit is an example of this construction. To understand this, you first need to understand the diagonal functor $$ \Delta: \mathcal{C} \to \mathcal{C}^J. $$ Here, $\mathcal{C}^J$ is the functor category of functors $F: J \to C$, with morphisms as natural transformations. This functor $\Delta$ takes each object $C$ to the functor $F_C: J \to C$, where for each $j \in J$ $$ F(j) = C. $$ So it sends it to a constant valued functor.

Now when we speak of a "colimit" in a category $\mathcal{C}$ it's with respect to some functor $F: J \to C$. This is an element of the functor category $\mathcal{C}^J$. So, we define a colimit to be a universal morphism from $F$ to $\Delta$. That is, $$ (\text{Colim }F, u: F \to \Delta(\text{Colim} F). $$ Note that $u$ is natural transformation; as I said earlier $\Delta$ sends objects to functors. So, you get the diagram

However, this isn't very intuitive. A better way to look at this is to realize that if you have a natural transformation $u: F \to \Delta(\text{Colim } F)$, then you have a family of morphisms. How so? For each object $i \in J$, our natural transformation should give us a morphism $$ u_i: F(i) \to \Delta(\text{Colim } F)(i). $$ But $\Delta(\text{Colim } F)$ is a constant valued functor. So this really becomes a family of morphisms $$ u_i: F(i) \to \text{Colim }F. $$ This is usually how people define colimits, but since you asked how they're related to universal constructions, this is how. Anyways, one way to imagine the above diagram is to picture the one below. This isn't exactly precise, but it's a good way to imagine the colimit.

Compare this diagram with the one in the beginning with the coproduct. If you understand that, then you can understand the concept of limits because limits are the same exact story, just with the arrows reversed.

Get vector $\vec{v}$ and on matrix $A$

[1 2 3][v1] = [1 2 3][1] = [6]
[2 2 1][v2]   [2 1 1][1]   [4]
       [v3]          [1]

v1+2v2+3v3=6 (1)
2v1+v2+v3=4  (2)

From (2): v3=4-v2-2v1

Plug into (1): v1+2v2+12-3v2-6v1=6
v2=-5v1+6

Plug back in (2): v3=4+5v1-6-2v1
v3=-2+3v1

One possible answer is:
Set v1=2
v2=-4
v3=4

All possible answers of the form:
[0 ] + [1 ]
[6 ]   [-5]x
[-2]   [3 ]

Best Answer

Related Solutions

Trying to understand the definition of a universal property

Get vector $\vec{v}$ and on matrix $A$

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