Subsection 11.1.4 (00XR): Precomposition and Postcomposition

Let $\mathcal{C}$ be a category and let $A,B,C,X\in \operatorname {\mathrm{Obj}}(\mathcal{C})$.

Let $f\colon A\to B$ and $g\colon B\to C$ be morphisms of $\mathcal{C}$.

1.
The precomposition function associated to $f$ is the function

\[ f^{*} \colon \operatorname {\mathrm{Hom}}_{\mathcal{C}}(B,X) \to \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,X) \]

defined by

\[ f^{*}(\phi ) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\phi \circ f \]

for each $\phi \in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(B,X)$.
2.
The postcomposition function associated to $g$ is the function

\[ g_{*} \colon \operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,B) \to \operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,C) \]

defined by

\[ g_{*}(\phi ) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g\circ \phi \]

for each $\phi \in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,B)$.

Let $A,B,C,D,X\in \operatorname {\mathrm{Obj}}(\mathcal{C})$.

1.
Interaction Between Precomposition and Postcomposition. Let $f\colon A\to B$ and $g\colon X\to Y$ be morphisms of $\mathcal{C}$. We have
2.
Interaction With Composition I. Let $f\colon A\to B$ and $g\colon B\to C$ be morphisms of $\mathcal{C}$. We have
3.
Interaction With Composition II. Let $f\colon A\to B$ and $g\colon B\to C$ be morphisms of $\mathcal{C}$. We have
4.
Interaction With Composition III. Let $f\colon X\to A$ and $g\colon C\to D$ be morphisms of $\mathcal{C}$. We have
5.
Interaction With Identities. We have

\begin{align*} \operatorname {\mathrm{id}}^{*}_{A} & = \operatorname {\mathrm{id}}_{\operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)},\\ (\operatorname {\mathrm{id}}_{B})_{*} & = \operatorname {\mathrm{id}}_{\operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)}. \end{align*}

Item 1: Interaction Between Precomposition and Postcomposition

For each $\phi \in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(B,X)$, we have

\begin{align*} [g_{*}\circ f^{*}](\phi ) & = g_{*}(\phi \circ f)\\ & = g\circ (\phi \circ f)\\ & = (g\circ \phi )\circ f\\ & = f^{*}(g\circ \phi )\\ & = [f^{*}\circ g_{*}](\phi ). \end{align*}

Thus $g_{*}\circ f^{*}=f^{*}\circ g_{*}$.

Item 2: Interaction With Composition I

$(g\circ f)_{*}=g_{*}\circ f_{*}$

For each $\phi \in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(X,A)$, we have

\begin{align*} (g\circ f)_{*}(\phi ) & = (g\circ f)\circ \phi \\ & = g\circ (f\circ \phi )\\ & = g\circ f_{*}(\phi )\\ & = g_{*}(f_{*}(\phi ))\\ & = [g_{*}\circ f_{*}](\phi ). \end{align*}

Thus $(g\circ f)_{*}=g_{*}\circ f_{*}$.

$(g\circ f)^{*}=g^{*}\circ f^{*}$

For each $\phi \in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(C,X)$, we have

\begin{align*} (g\circ f)^{*}(\phi ) & = \phi \circ (g\circ f)\\ & = (\phi \circ g)\circ f\\ & = (g^{*}(\phi ))\circ f\\ & = f^{*}(g^{*}(\phi ))\\ & = [f^{*}\circ g^{*}](\phi ). \end{align*}

Thus $(g\circ f)^{*}=g^{*}\circ f^{*}$.

Item 3: Interaction With Composition II

It suffices to show the equalities of the maps on $\star \in \mathrm{pt}$. We have

\begin{align*} [g\circ f](\star ) & = g\circ f\\ & = g_{*}(f)\\ & = g_{*}([f](\star ))\\ & = (g_{*}\circ [f])(\star ), \end{align*}

and

\begin{align*} [g\circ f](\star ) & = g\circ f\\ & = f^{*}(g)\\ & = f^{*}([g](\star ))\\ & = (f^{*}\circ [g])(\star ).\end{align*}

Thus $[g\circ f]=g_{*}\circ [f]$ and $[g\circ f]=f^{*}\circ [g]$.

Item 4: Interaction With Composition III

$f^{*}\circ \mathord {\circ ^{\mathcal{C}}_{A,B,C}}=\mathord {\circ ^{\mathcal{C}}_{X,B,C}}\circ (\mathsf{id}\times f^{*})$

For each $(\psi ,\phi )\in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(B,C)\times \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)$, we have

\begin{align*} [f^{*}\circ \mathord {\circ ^{\mathcal{C}}_{A,B,C}}](\psi ,\phi ) & = f^{*}(\psi \circ \phi )\\ & = (\psi \circ \phi )\circ f\\ & = \psi \circ (\phi \circ f)\\ & = \mathord {\circ ^{\mathcal{C}}_{X,B,C}}(\psi ,\phi \circ f)\\ & = \mathord {\circ ^{\mathcal{C}}_{X,B,C}}(\psi ,f^{*}(\phi ))\\ & = [\mathord {\circ ^{\mathcal{C}}_{X,B,C}}\circ (\mathsf{id}\times f^{*})](\psi ,\phi ). \end{align*}

Thus $f^{*}\circ \mathord {\circ ^{\mathcal{C}}_{A,B,C}}=\mathord {\circ ^{\mathcal{C}}_{X,B,C}}\circ (\mathsf{id}\times f^{*})$.

$g_{*}\circ \mathord {\circ ^{\mathcal{C}}_{A,B,C}}=\mathord {\circ ^{\mathcal{C}}_{A,B,D}}\circ (g_{*}\times \mathsf{id})$

For each $(\psi ,\phi )\in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(B,C)\times \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)$, we have

\begin{align*} [g_{*}\circ \mathord {\circ ^{\mathcal{C}}_{A,B,C}}](\psi ,\phi ) & = g_{*}(\psi \circ \phi )\\ & = g\circ (\psi \circ \phi )\\ & = (g\circ \psi )\circ \phi \\ & = \mathord {\circ ^{\mathcal{C}}_{A,B,D}}(g\circ \psi ,\phi )\\ & = \mathord {\circ ^{\mathcal{C}}_{A,B,D}}(g_{*}(\psi ),\phi )\\ & = [\mathord {\circ ^{\mathcal{C}}_{A,B,D}}\circ (g_{*}\times \mathsf{id})](\psi ,\phi ). \end{align*}

Thus $g_{*}\circ \mathord {\circ ^{\mathcal{C}}_{A,B,C}}=\mathord {\circ ^{\mathcal{C}}_{A,B,D}}\circ (g_{*}\times \mathsf{id})$.

Item 5: Interaction With Identities

We have

\begin{align*} \operatorname {\mathrm{id}}^{*}_{A}(\phi ) & = \phi \circ \operatorname {\mathrm{id}}_{A}\\ & = \phi \\ & = \operatorname {\mathrm{id}}_{\operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)}(\phi )\end{align*}

and

\begin{align*} (\operatorname {\mathrm{id}}_{B})_{*}(\phi ) & = \operatorname {\mathrm{id}}_{B}\circ \phi \\ & = \phi \\ & = \operatorname {\mathrm{id}}_{\operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)}(\phi )\end{align*}

for each $\phi \in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)$.

The Clowder Project

11.1.4 Precomposition and Postcomposition