How to Find limits and co-limits of diagrams over Vector?

Question

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.

Answer 1

$\require{AMScd}\newcommand\mat[1]{\begin{bmatrix}#1\end{bmatrix}}$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.

Answer 2

I think you can start by computing in the simpler cases. First ask what is the (categorical) product and the (categorical) sum for finite dimensional vector spaces. With the help of the hint you have (look at usual product and usual sum of vector spaces).

From there instead of of trying to guess directly what the limit and colimit you want are, you can try computing explicitly first what the pullback and pushout look like in the general case (think about what happens in the category set for a guideline!). Once you have done that, you can decompose the (co)limit you want into successive pullback/pushout, and use the fact that two pullback squares are again a pullback square (idem for pushout). This should let you do all the computation explicitly, by breaking down to easier cases.