Morphisms as Objects, Comm. Diagrams etc.

Question

Let $C$ be a category. Denote by $Ar(C)$ the following category: an object in $Ar(C)$ is a morphism $X_1 \rightarrow X_2$ in $C$. A morphism in $Ar(C)$ from $X_1 \rightarrow X_2$ to $Y_1 \rightarrow Y_2$ is a commutative diagram.

1) Show that $Ar(C)$ is a category.

2) Show that $F:Ar(Ab) \rightarrow Ar(Ab)$ with $F(X_1 \rightarrow X_2) = (\mathrm{ker}(f) \rightarrow X_1)$ defines a functor.

1) If $f,g$ are objects, $\mathrm{Hom}(f,g)$ is basically a pair of morphisms $(\varphi,\gamma)$ such that $g \varphi=\gamma f$. Let $(\varphi,\gamma) \in \mathrm{Hom}(f,g)$ and $(\gamma',\chi) \in \mathrm{Hom}(g,h)$. I'm unsure what needs to be done with composition. We need to be able to compose $((\varphi,\gamma),(\gamma',\chi))$ so that it lies in $\mathrm{Hom}(f,h)$ and is associative. Define $(\gamma',\chi) \circ (\varphi,\gamma)=(\gamma' \varphi,\chi \gamma)$. This is associative, but I don't know if it's such that $h \gamma' \varphi = \chi \gamma f$? Certainly they're both maps $X_1 \rightarrow Z_2$ ($h:Z_1 \rightarrow Z_2$), and I don't know how to convince myself whether it is or not?

Secondly for this question, Rotman writes of categories:

for each object $A$, there is an identity morphism $1_A \in \mathrm{Hom}(A,B)$ such that $f1_A=f$ and $1_B f=f$ for all $f:A \rightarrow B$.

This confuses me. Let $f:X_1 \rightarrow X_2$, $g: Y_1 \rightarrow Y_2$ and $(\varphi,\gamma) \in \mathrm{Hom}(f,g)$. So I have to show that there is $1_f=(\alpha,\beta)\in \mathrm{Hom}(f,f)$ such that $(\varphi,\gamma)(\alpha,\beta)=(\varphi \alpha,\gamma \beta)=(\varphi,\gamma)$ and $1_g (\varphi,\gamma)=(\varphi,\gamma)$. Well, letting $\alpha$ be the identity on $X_1$ and so forth seems to do the trick, but I don't understand, from Rotman, why $1_B$, or in my case $1_g$, would have to exist. Seems more reasonable to show that both exist. Perhaps that's just a minor issue...

2) I'm not sure here how the functor works on morphisms. Let $(\varphi,\gamma) \in \mathrm{Hom}(f,g)$. We must have $F(\varphi,\gamma):F(f) \rightarrow F(g)$, or in words, $(\varphi,\gamma)$ must be mapped to a pair making $\mathrm{ker}(f) \rightarrow X_1$ and $\mathrm{ker}(g) \rightarrow Y_1$ commute. I see two ways here; either introduce some map $\mathrm{ker}(f) \rightarrow \mathrm{ker}(g)$ and restrict $\varphi$ somehow, neither of which I'd have a clue on how to do, or look at how $F$ works on $\varphi,\gamma$ separately. In the latter case, $F(\varphi):\mathrm{ker}(\varphi) \rightarrow X_1$ and $F(\gamma):\mathrm{ker}(\gamma) \rightarrow X_2$. The first might make sense, but $F(\gamma)$ makes no sense to me in trying to make the new square diagram commute. Hints on how to show that $F$ is a functor?

Answer 1

This should really probably be two different questions so I'll only help with the first here. Your idea for how composition in $Ar(C)$ works is correct. It's perhaps easier to think about what's going on in terms of commutative diagrams. If $(\varphi,\gamma)\in\mbox{Hom}(f,g)$ and $(\varphi',\gamma')\in\mbox{Hom}(g,h)$, then the following diagrams commute:

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} X_1 & \ra{f} & X_2 \\ \da{\varphi} & & \da{\gamma} \\ Y_1 & \ras{g} & Y_2 \\ \end{array}\quad\&\quad \begin{array}{c} Y_1 & \ra{g} & Y_2 \\ \da{\varphi'} & & \da{\gamma'} \\ Z_1 & \ras{h} & Z_2 \\ \end{array} $$ and their composition is given by just 'stacking' the two diagrams vertically like $$\begin{array}{c} X_1 & \ra{f} & X_2 \\ \da{\varphi} & & \da{\gamma} \\ Y_1 & \ras{g} & Y_2 \\ \da{\varphi'} & & \da{\gamma'} \\ Z_1 & \ras{h} & Z_2 \\ \end{array}$$ giving an arrow from $f$ to $h$ which we write as $(\varphi\circ\varphi', \gamma\circ\gamma')$. The above diagram commutes because the individual squares commute but we can prove this 'algebraically' by noting that $\gamma\circ f=g\circ \varphi$ and $\gamma'\circ g= h\circ \varphi'$ implies that $(\gamma'\circ \gamma) \circ f=\gamma'\circ (\gamma \circ f)= \gamma'\circ (g\circ \varphi) = (\gamma'\circ g)\circ \varphi= (h\circ \varphi') \circ \varphi = h\circ (\varphi' \circ \varphi)$ as required.

The associativity of composition of commuting squares is given by the associaitivity of arrows in $C$ which I won't bother writing down (if you're still unclear feel free to ask), as it follows very quickly by just noting that sticking squares together like $(\square \square) \square$ is the same as sticking them together like $\square (\square\square)$.

The identity for $f\colon X_1\rightarrow X_2$ is given by the pair $\mbox{Id}_f = (\mbox{Id}_{X_1},\mbox{Id}_{X_2})$ as $\mbox{Id}_{X_2}\circ f = f = f\circ \mbox{Id}_{X_1}$ so it gives a commuting square. We see it's the identity because if $(\varphi,\gamma)\colon f\rightarrow g$, then $(\mbox{Id}_{X_1},\mbox{Id}_{X_2})\circ (\varphi,\gamma) = (\mbox{Id}_{X_1}\circ \varphi,\mbox{Id}_{X_2}\circ\gamma) = (\varphi,\gamma)$ and similarly if $(\varphi',\gamma')\colon h\rightarrow f$ we would get $(\varphi',\gamma') \circ (\mbox{Id}_{X_1}, \mbox{Id}_{X_2}) = (\varphi',\gamma')$. So, $\mbox{Id}_f$ is the identity element on $f$ in $Ar(C)$ as required.