There is a step in J.F.Adams book, Infinite Loop Spaces, which I don't quite understand. Here is the whole extract:
Let $W$ be a further space (not sure what 'further' means, seems unnecessary), with base-point $w_0$. Denote $X^I$ by the set of all functions from $I$ to $X$. Then maps
$$f:W\rightarrow X^I$$
are in 1-1 correspondence with maps
$$g:W\times I\rightarrow X$$
in the following way:
$$(fw)(t)=g(w,t), \hspace{10mm}(w\in W,t\in I)$$
If we take account of the base-points, we find that maps
$$f:W,w_0\rightarrow\Omega X,\omega_0$$
are in 1-1 correspondence with maps
$$g:\Sigma W,\sigma_0\rightarrow X,x_0$$
Here $\Sigma W$ is the quotient space obtained from $W\times I$ by identifying the subspace $(W\times0)\cup(w_0\times I)\cup (W\times 1)$ to a single point, which becomes the base-point $\sigma_0$ in $\Sigma W$.
Question: I sort of understand why we need to identify $(W\times0)\cup (W\times 1)$ - this is because we need to have $g(w,0)=g(w,1)$ for all $g$ (am I right?). However I don't really see why we need to quotient out $(w_0\times I)$. It would be nice if someone could provide an explicit example for this. Thanks!
It is basically because $f$ maps the base-point of $W$ to the base-point of $\Omega X$, which is the constant loop.
Identifying $w_0\times I$ makes sure that the loop $w_0$ corresponds to is actually constant.
For an explicit example just consider $W=\{w_0\}$. The cardinality of $\hom_*(W,\Omega X)$ is $1$, thus $\hom_*(\Sigma W,X)$ must have cardinality $1$ as well, for any $X$. This can clearly only be true if $\Sigma W = \{\sigma_0\}$ rather than $\{w_0\}\times I$.