# Commutativity of triangle consisting of morphism and two adjuncts

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. :)

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.

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}