TITLE: Why isn't this homotopy pullback a point?
QUESTION [7 upvotes]: Consider any point in a topological space $X$ and consider the homotopy pullback of the diagram $* \to X \leftarrow *$. Why is said pullback the loop space of $X$ instead of just a point? 
If we have any other space $P$ then we have precisely one map $P\to *$, so we can only consider one diagram involving $P$,
$$\require{AMScd} \begin{CD}
    P @>{}>> *\\ @V{}VV @VV{}V\\
    * @>>{}> X
    \end{CD}$$
and it clearly factors through the point via the only possible homotopy class of map from $P$ to $*$, the constant map. So it appears that the point space fulfills the universal property that defines the homotopy pullback. I understand that if you consider the standard construction of the homotopy pullback, you immediately obtain $\Omega X$, but I fail to see why the point does not fulfill the universal property defining the pullback.

REPLY [2 votes]: Here is an alternative perspective to the one laid out by Najib Idrissi. 
In any category $C$ with weak equivalences, you can define the homotopy limit over a functor $F:J\to C$ as the value of right derived functor $\mathbb{R}{\rm lim}$ of the limit functor, evaluated on $F$ 
In good cases, you can calculate this right derived functor explicitly in terms of "weighted" limits, or ends (this is discussed a bit, for example, at the nlab page https://ncatlab.org/nlab/show/homotopy+limit under the section "local definition"). These weighted limits make good sense of what it means to be a homotopy commutative cone over a diagram. Such weighted limits are sometimes called $\infty$-limits. 
In the language of $\infty$-categories, there is a precise meaning of what it means to be a homotopy commutative (more precisely homotopy coherent) cone over a diagram and one can make sense of what it means to be a universal such cone.
Going back to your question, homotopy limits in the category of $CW$-complexes is one of these good cases. An $\infty$-cone over a diagram of the form $\bullet \to \bullet \leftarrow \bullet$ can be identified with a homotopy commutative diagram 
$$\require{AMScd}
\begin{CD}
 X @>>> \ast \\
@VVV H @VyVV \\
\ast @>y>> Y
\end{CD}$$
where we have an explicit choice of homotopy $H:X\times I\to Y$ with $H(x,0)=y$ and $H(x,1)=y$ for all $x$. Now let's say $X$ is the universal such cone over the diagram. Consider the homotopy commutative diagram 
$$\require{AMScd}
\begin{CD}
 \ast @>>> \ast \\
@VVV \gamma @VyVV \\
\ast @>y>> Y
\end{CD}$$
The choice of homotopy gives rise to a map $\gamma:I\to Y$ such that $H(\ast,0)=y=H(\ast,1)$; i.e. a loop in $Y$. By the homotopy coherent universal property, there is an induced map $\ast\to X$, which is not unique, but is unique up to a contractible space of higher homotopies filling the diagram. So for each loop in $Y$, we have a corresponding point $x:\ast\to X$, which is not uniquely determined by $\gamma$, but is up to some contractible space. 
You can continue this process and consider the homotopy commutative cones 
$$\require{AMScd}
\begin{CD}
 S^n @>>> \ast \\
@VVV \gamma_n @VyVV \\
\ast @>y>> Y
\end{CD}$$
where $\gamma_n:S^n\times [0,1]\to Y$ satisfies $\gamma_n(x,0)=y=\gamma_n(x,1)$. 
Let's take a step back and see what we have just described. For each $n$, we have a well defined function $f:\hom(S^n,X)\to \hom(S^n,\Omega_yY)$, coming from the canonical homotopy $H$ which fills the universal diagram. We don't have a well defined map backwards but after quotienting out by the contractible space of possible fillers in the diagram, we do get a well defined map on homotopy classes, this gives a map $\pi_n(\Omega_yY)\to \pi_n(X)$, which is easily checked to be an isomorphism (with inverse $f_*$).