How can I prove that $[\sum X, Y] \cong [X, \Omega Y]$?
Could anyone please mention a reference for me that contained a detailed proof of this or show me the detailed proof of this? My professor gave it to us during the lecture without explaining the details of the proof.
There are two things to check. That the bijection makes sense, and that it preseves the group structure. Unfortunately the answer suggested in the comments does neither of these things.
For pointed spaces $X,Y$ let $Top_*(X,Y)$ denote the set of basepoint preserving maps $X\rightarrow Y$, and $Map_*(X,Y)$ this set given the compact-open topology. If $Z$ is another pointed space and $Z$ is locally compact, then there is a bijection
$$Top_*(X\wedge Z,Y)\cong Top_*(X,Map_*(Z,Y))\qquad\qquad\qquad (\ast)$$
which is induced by sending a map $f:X\wedge Z\rightarrow Y$ to the map $\hat f:X\rightarrow Map_*(Z,Y)$ given by
$$\hat f(x)(z)=f(x\wedge z),\qquad x\in X,z\in Z.$$
I'll let you check that $\hat f$ is continuous and write down the inverse to the assignment $f\mapsto \hat f$.
The bijection $(\ast)$ preserves the relation of pointed homotopy. Indeed, $I_+=I\sqcup\{+\}$ is locally compact, so if $Z$ is also, then there is a homeomorphism
$$(X\wedge Z)\wedge I_+\cong (X\wedge I_+)\wedge Z$$
which is natural in $X$. Applying the previous observation we have bijections
$$Top_*((X\wedge Z)\wedge I_+,Y)\cong Top_*((X\wedge I_+)\wedge Z,Y)\cong Top_*(X\wedge I_+,Map_*(Z,Y))$$
which are natural in both $X$ and $Y$. This is the statement that there is a one-to-one correspondence between pointed homotopies of maps $X\wedge Z\rightarrow Y$ and pointed homotopies of maps $X\rightarrow Map_*(Z,Y)$.
Thus if $X,Y$ are any pointed spaces and $Z$ is locally compact, then there is a bijection
$$[X\wedge Z,Y]\cong [X,Map_*(Y,Z)]$$
where the square brackets denote pointed homotopy classes.
Since $S^1$ is locally compact, we can set $Z=S^1$ in the above to get $$[\Sigma X,Y]\cong [X,\Omega Y]$$ naturally in $X,Y$. Here and below I will use $\Sigma X=X\wedge S^1$ and $\Omega Y=Map_*(S^1,Y)$ as definitions.
It remains to check the group structure. The group structure on $[\Sigma X,Y]$ is induced by the pinch map $c$ on the suspension. Given $f,g:\Sigma X\rightarrow Y$, the product $f\cdot g$ is defined as the composition $$f\cdot g:\Sigma X\xrightarrow{c} \Sigma X\vee\Sigma X\xrightarrow{f\vee g}Y\vee Y\xrightarrow{\nabla}Y$$ It is a standard result that the homotopy class of $f\cdot g$ depends only on the homotopy classes of $f$ and $g$, so the product descends to $[\Sigma X,Y]$. Similarly, in $[X,\Omega Y]$, the group structure is induced by the loop multiplication $m$. If $h,k:X\rightarrow \Omega Y$ are given, their product is defined as $$h\star k:X\xrightarrow{\Delta}X\times X\xrightarrow{h\times k}\Omega Y\times \Omega Y\xrightarrow{m}\Omega Y.$$
Notice that with regards to the adjunction $(\ast)$ we have
$$\widehat{\alpha\circ \beta}=\Omega \alpha\circ\hat \beta$$
whenever $X\xrightarrow \beta Y\xrightarrow \alpha Z$ are composable. Using this I'll let you check the fairly obvious string of homotopies
$$\widehat{f\cdot g}=\Omega\nabla\circ\Omega(f\vee g)\circ\hat c\simeq m\circ(\Omega f\times \Omega g)\circ (\widehat{id_X}\times \widehat {id_X})\circ\Delta=\hat f\star\hat g.$$
Since the constant map is the unit in both groups $[\Sigma X,Y]$ and $[X,\Omega Y]$, and the multiplicative adjunction $(\ast)$ sends one to the other we see that
$$[\Sigma X,Y]\cong [X,\Omega Y]$$
is an isomorphism of groups.