# Does a function space construction always decrease the connectivity of a space?

This question is kind of dual to this question, where I asked if smashing with a space $Y$ always increases (which means ''$\geq$'') the connectivity of a space $X$ and the answer was "yes".

The connectivity of a pointed space $X$ is the maximal number $\operatorname{con}(X)$ such that $\pi_i(X)=0$ for all $0\leq i\leq\operatorname{con}(X)$.

Let $X$ and $Y$ be pointed CW complexes and equip the set $\operatorname{Hom_\bullet} (Y,X)$ of pointed continuous maps with the compact-open topology.

Is it true that $\operatorname{con}(\operatorname{Hom_\bullet} (Y,X))\leq \operatorname{con}(X)$ if $X$ is connected?

This is particullary interesting when $Y=S^1$ where $\operatorname{Hom_\bullet} (S^1,X)=\Omega(X)$ but then $\pi_n(\Omega(X))=\pi_{n+1}(X)$ and the statement is true in this case.

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Isn't the answer clearly no? If $Y$ is a pointed interval, $Hom_*(Y,X)$ is the path-space of $X$, $P(X)$, which is contractible. But $X$ isn't usually contractible. – Ryan Budney Feb 7 '12 at 16:30
If something like this is true, then you should be able to verify it using the spectral sequence for a mapping space (obtained either by a skeletal filtration of the source or a Postnikov power for the target) -- this is really a special case of the Bousfield-Kan spectral sequence. This (the special case) is in Mosher & Tangora's book, for instance. – Aaron Mazel-Gee Oct 22 '12 at 2:24

But if $\dim Y\le\operatorname{con}X$ and $X$ is connected, then
• $\operatorname{map}(Y,X)$ is connected;
• $\operatorname{con}\operatorname{map}(Y,X)\ge\operatorname{con}X-\dim Y$ (because $\pi_i\operatorname{map}(Y,X)=\pi_0\operatorname{map}(Y,\Omega^i X)$, for example).
And I don't think there is an easy estimate in the direction you want. For example, $\operatorname{con}(S^1\times S^1)=0$ but $\operatorname{con}\operatorname{map}(S^2,S^1\times S^1)=+\infty$ (maps from $S^2$ can be lifted to the universal cover -- which is contractible).