A Seven-Term Adjoint Chain in the Arrow Category
A direct proof with every relevant commutative square drawn explicitly.
Let $\mathcal A$ be an abelian category. Its arrow category $\mathcal A^{[1]}$ is the functor category associated with
$$ [1]=(0\longrightarrow 1). $$It is also often denoted by $\mathcal A^2$. An object of $\mathcal A^{[1]}$ is simply a morphism
$$ f:X_0\longrightarrow X_1 $$in $\mathcal A$.
A morphism $$ (a,b):f\longrightarrow g $$ from $f:X_0\to X_1$ to $g:Y_0\to Y_1$ is a commutative square, equivalently a pair of morphisms
$$ a:X_0\longrightarrow Y_0, \qquad b:X_1\longrightarrow Y_1 $$satisfying $ga=bf$. In diagrammatic form:
A morphism in the arrow category.
$$ \begin{CD} X_0 @>{a}>> Y_0 \\ @V{f}VV @VV{g}V \\ X_1 @>{b}>> Y_1 \end{CD} $$Here
$$ \operatorname{ev}_i(X_0\xrightarrow{f}X_1)=X_i, \qquad L_1(A)=(0\to A), \qquad H(A)=(A\xrightarrow{\operatorname{id}_A}A), \qquad L_0(A)=(A\to 0). $$1. The complete statement
For every arrow $f:X_0\to X_1$ and every object $A\in\mathcal A$, there are natural bijections
$$ \operatorname{Hom}_{\mathcal A}(\operatorname{coker}f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_1A), $$ $$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_1A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\operatorname{ev}_1f), $$ $$ \operatorname{Hom}_{\mathcal A}(\operatorname{ev}_1f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,HA), $$ $$ \operatorname{Hom}_{\mathcal A^{[1]}}(HA,f) \cong \operatorname{Hom}_{\mathcal A}(A,\operatorname{ev}_0f), $$ $$ \operatorname{Hom}_{\mathcal A}(\operatorname{ev}_0f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_0A), $$and
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_0A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\ker f). $$2. The adjunction $\operatorname{coker}\dashv L_1$
We prove
$$ \operatorname{Hom}_{\mathcal A}(\operatorname{coker}f,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_1A). $$Let $q_f:X_1\to\operatorname{coker}f$ be the cokernel morphism. A morphism $f\to L_1A$ is a commutative square
A morphism $f\to L_1A$.
$$ \begin{CD} X_0 @>{0}>> 0 \\ @V{f}VV @VV{0}V \\ X_1 @>{b}>> A \end{CD} $$Since the top arrow is necessarily zero, commutativity is exactly the condition $bf=0$. By the universal property of the cokernel, this is equivalent to the existence of a unique morphism
$$ \overline b:\operatorname{coker}f\longrightarrow A $$such that $b=\overline b\,q_f$. Hence
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_1A) \cong \operatorname{Hom}_{\mathcal A}(\operatorname{coker}f,A). $$So
$$ \boxed{\operatorname{coker}\dashv L_1}. $$3. The adjunction $L_1\dashv\operatorname{ev}_1$
We prove
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_1A,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_1). $$A morphism $L_1A\to f$ is a commutative square
A morphism $L_1A\to f$.
$$ \begin{CD} 0 @>{0}>> X_0 \\ @V{0}VV @VV{f}V \\ A @>{b}>> X_1 \end{CD} $$The commutativity condition $f\circ 0=b\circ 0$ is automatic, so the square is determined uniquely by $b:A\to X_1$. Therefore
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_1A,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_1) = \operatorname{Hom}_{\mathcal A}(A,\operatorname{ev}_1f). $$Thus
$$ \boxed{L_1\dashv\operatorname{ev}_1}. $$4. The adjunction $\operatorname{ev}_1\dashv H$
We prove
$$ \operatorname{Hom}_{\mathcal A}(X_1,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,HA). $$A morphism $f\to HA$ is a commutative square
A morphism $f\to HA$.
$$ \begin{CD} X_0 @>{a}>> A \\ @V{f}VV @VV{\operatorname{id}_A}V \\ X_1 @>{b}>> A \end{CD} $$Commutativity means $\operatorname{id}_A\circ a=bf$, hence $a=bf$. So once $b:X_1\to A$ is chosen, the top arrow is forced. Therefore
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(f,HA) \cong \operatorname{Hom}_{\mathcal A}(X_1,A). $$Hence
$$ \boxed{\operatorname{ev}_1\dashv H}. $$5. The adjunction $H\dashv\operatorname{ev}_0$
We prove
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(HA,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_0). $$A morphism $HA\to f$ is a commutative square
A morphism $HA\to f$.
$$ \begin{CD} A @>{a}>> X_0 \\ @V{\operatorname{id}_A}VV @VV{f}V \\ A @>{b}>> X_1 \end{CD} $$Commutativity means $fa=b\,\operatorname{id}_A$, hence $b=fa$. So once $a:A\to X_0$ is chosen, the bottom arrow is forced. Therefore
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(HA,f) \cong \operatorname{Hom}_{\mathcal A}(A,X_0). $$Hence
$$ \boxed{H\dashv\operatorname{ev}_0}. $$6. The adjunction $\operatorname{ev}_0\dashv L_0$
We prove
$$ \operatorname{Hom}_{\mathcal A}(X_0,A) \cong \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_0A). $$A morphism $f\to L_0A$ is a commutative square
A morphism $f\to L_0A$.
$$ \begin{CD} X_0 @>{a}>> A \\ @V{f}VV @VV{0}V \\ X_1 @>{0}>> 0 \end{CD} $$The commutativity condition $0\circ a=0\circ f$ is automatic, so such a square is determined uniquely by $a:X_0\to A$. Therefore
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(f,L_0A) \cong \operatorname{Hom}_{\mathcal A}(X_0,A) = \operatorname{Hom}_{\mathcal A}(\operatorname{ev}_0f,A). $$Hence
$$ \boxed{\operatorname{ev}_0\dashv L_0}. $$7. The adjunction $L_0\dashv\ker$
We prove
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_0A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\ker f). $$Let $i_f:\ker f\to X_0$ be the kernel morphism. A morphism $L_0A\to f$ is a commutative square
A morphism $L_0A\to f$.
$$ \begin{CD} A @>{a}>> X_0 \\ @V{0}VV @VV{f}V \\ 0 @>{0}>> X_1 \end{CD} $$Commutativity means $fa=0$. By the universal property of the kernel, this is equivalent to the existence of a unique morphism
$$ \overline a:A\longrightarrow\ker f $$such that $a=i_f\overline a$. Hence
$$ \operatorname{Hom}_{\mathcal A^{[1]}}(L_0A,f) \cong \operatorname{Hom}_{\mathcal A}(A,\ker f). $$Therefore
$$ \boxed{L_0\dashv\ker}. $$8. Summary of all six special squares
For convenience, here are the six special commuting squares that appear in the proof:
9. Conceptual explanation via Kan extensions
The middle five functors come from restriction along the endpoint inclusions
$$ i_0:\{0\}\hookrightarrow [1], \qquad i_1:\{1\}\hookrightarrow [1]. $$Restriction gives
$$ i_0^*=\operatorname{ev}_0, \qquad i_1^*=\operatorname{ev}_1. $$For $i_1$ one gets
$$ \operatorname{Lan}_{i_1}=L_1, \qquad \operatorname{Ran}_{i_1}=H, $$hence
$$ L_1\dashv\operatorname{ev}_1\dashv H. $$For $i_0$ one gets
$$ \operatorname{Lan}_{i_0}=H, \qquad \operatorname{Ran}_{i_0}=L_0, $$hence
$$ H\dashv\operatorname{ev}_0\dashv L_0. $$The overlap $H=\operatorname{Ran}_{i_1}=\operatorname{Lan}_{i_0}$ glues the two triples into the five-term chain
$$ L_1\dashv\operatorname{ev}_1\dashv H\dashv\operatorname{ev}_0\dashv L_0. $$Finally, kernel and cokernel extend the chain at the ends:
$$ \operatorname{coker}\dashv L_1, \qquad L_0\dashv\ker. $$
