In case this question appeared here I apologize for the preposterous usage of computer space and attention of participants. I searched over there, for a similar or the same question for an answer but resigned. :)


I am reading: 


S. B. Niefield, Cartesianness, topological spaces, uniform spaces, and affine schemes, *J. Pure Appl. Alg.* **23**, 1982, 147–167, where the author extends the Day-Kelley characterization of exponentiable objects in $Top$ to slice categories $Top/X$.


Let $A$ be a category with finite limits and $T$ is an object of $A$. It is clear that a product $X\times T$ becomes an object over $T$ via the projection $\pi\_2 : X\times T\rightarrow T$. This induces a functor $T^\*:A\rightarrow A/T$ which is clearly right adjoint of the forgetful functor $\Sigma\_T :A/T\rightarrow A$.


Now the ***Proposition 1.1*** states: A functor $F:B\to A/T$ has a right adjoint if and only if $\Sigma\_T \circ F$ has a right adjoint.


To prove that if $\Sigma\_T \circ F$ has a right adjoint then $F$ has a right adjoint the author argues: 
Suppose $\Sigma\_T \circ F\dashv G'$, For $X$ object of $B$, let $\sigma\_X:X\to G'T$ be the right adjunct of $FX$ considered as morphism $\Sigma\_T(FX)\to T$. Then, if $f:X\to X'$ is a morphism of $B$, the diagram 
$\require{AMScd}$
\begin{CD}
 X @>f>> X' \\
 @V \sigma\_X VV= @VV \sigma\_{X'} V\\
 G'T @>>id\_{G'T}> G'T
\end{CD}
commutes. 


**I stuck on this: WHY THE DIAGRAM COMMUTES?**


Then using naturality (since the commutation of the above diagram means that $\sigma $ is a natural transformation) the author just builds the right adjoint $G$ for $F$ as equalizer of


$G'p\_Y :G'Y\to G'T$ 


and 


$\sigma \_{G'Y}:G'Y\to G'T$ 


where $p\_Y: Y\to T$ is an object of $A/T$.


My Question is why the diagram commutes. 
Thanks in advance.



Commutativity of triangle consisting of morphism and two adjuncts
If $f:X \to X'$ is a morphism, then we have an induced morphism $F(f) : F(X) \to F(X')$. By definition of $A/T$, this is a commutative diagram


$\require{AMScd}$
\begin{CD}
 \Sigma\_T(F(X)) @>\Sigma\_T(F(f))>> \Sigma\_T(F(X')) \\
 @V VV @VV V\\
 T @>\mathrm{id}>> T 
\end{CD}


This gives the desired diagram


\begin{CD}
 X @>f>> X' \\
 @V \sigma\_X VV @VV \sigma\_{X'} V\\
 G'T @>>\mathrm{id}> G'T
\end{CD}


because the bijection in the adjunction is natural.


