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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!

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It looks like it means you need the path as well as the endpoints. $W$ represents a map object and so $(w_0\times I)$ is a single map from $I$ to $X$. –  user69810 Apr 19 '13 at 0:32
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I am not sure what the base point is but a second glance looks like the quotient space is obtained by taking all paths in $X$ and mapping them to the base point which looks like a single path in $X$. If this isn't correct I hope it's at least helpful. –  user69810 Apr 19 '13 at 0:44
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1 Answer

up vote 1 down vote accepted

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 $\{w_0\}\times I$.

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Thanks for your answer! Just one question - for your example are you assuming the fact that the sets $[W,\Omega X]$ and $[\Sigma W,X]$ have the same cardinality? If yes then the example is sort of circular - the above extract was to show this fact. If not - I don't see why it's obvious. –  Steve Apr 19 '13 at 1:54
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The above example shows that $[W,\Omega X]$ can only have the same cardinality as $[\Sigma W,X]$ if $\Sigma$ includes quotienting by $w_0\times I$. So if you want to show that $[W,\Omega X]$ and $[\Sigma W,X]$ have the same cardinality you have to define $\Sigma$ the usual way. –  Abel Apr 19 '13 at 11:55
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