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$[X,X]$ is the collection of homotopy equivalences from $X$ to itself, with $X$ a $CW$ complex.

To start, suppose that $X$ is a $1$-dimensional $CW$ complex. As usual, let $\mathcal{M}(X):=Homeo^+(X)/Homeo_0$ be the mapping class group.

For surfaces $\Sigma_g$ with $g \geq 1$, we know that a homotopy equivalence of $X$ with itself is homotopic to some homeomorphism, but there are spaces for which this fails, such as the wedge of two circles. In general, I want to measure how far $[X,X]$ is from $\mathcal{M}(X)$

I know that for any eilenberg-maclane space, $[X,X] \cong Out(\pi_1(X))$, so for any rose, $[X,X] \cong Out(F_n)$, whereas the mapping class group is finite.

$Q1:$ Is $\mathcal{M}(X)$ normal in $[X,X]$, and if so, what can be said of the quotient? Is there a class of spaces that makes $\mathcal{M}(X)$ normal?

$Q2:$ Are there references for this sort of question?

$Q3:$ how about for higher dimensional $CW$ complexes?

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    $\begingroup$ Pardon my ignorance: why is $[X,X]$ a group? Is $X$ any manifold, a surface, or a $K(G,1)$? (The last case I understand.) $\endgroup$ – Randall Nov 3 '17 at 18:03
  • $\begingroup$ Excuse me, homotopy equivalences $\endgroup$ – Andres Mejia Nov 3 '17 at 18:05
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    $\begingroup$ The case of a 1-dimensional CW complex $X$ can be reduced to a discrete situation: the mapping class group of $X$ is the group of isomorphisms of $X$ thought of as a graph. To be precise, let $\mathcal{E}$ be the set of oriented edges; let $r : \mathcal{E} \to \mathcal{E}$ be the involution that reverses orientation; let $V$ be the set of vertices; and let $i : \mathcal{E} \to V$ be the map which assigns to each oriented edge its initial vertex... $\endgroup$ – Lee Mosher Nov 4 '17 at 23:02
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    $\begingroup$ ...Then the mapping class group of $X$ is isomorphic to the set of permutations $\sigma$ of the disjoint union $\mathcal{E} \coprod V$ such that $\sigma(\mathcal{E})=\mathcal{E}$, $\sigma \circ r = r \circ \sigma$, and $\sigma \circ i = i \circ \sigma$. $\endgroup$ – Lee Mosher Nov 4 '17 at 23:02
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    $\begingroup$ The proof of this general version reduces to the 1-dimensional versiton of a theorem of Alexander: every homeomorphism of $[0,1]$ that fixes $0$ and $1$ os isotopic to the identity. $\endgroup$ – Lee Mosher Nov 4 '17 at 23:04
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I'll turn my comments about the 1-dimensional case into a partial answer addressing only that case, and I'll add a proof that normality fails.

When $X$ is a 1-dimensional CW complex, also known to topologists as a "graph", then calculation of the mapping class group $\mathcal{M}(X)$ can be reduced to a discrete problem. Let $\mathcal{E}$ be the set of oriented edges; let $r : \mathcal{E} \to \mathcal{E}$ be the involution that reverses orientation; let $V$ be the set of vertices, and let $i : \mathcal{E} \to V$ be the map which assigns to each oriented edge its initial vertex. Then $\mathcal{M}$ is isomorphic to the group of permutations $\sigma$ of the disjoint union $\mathcal{E} \coprod V$ such that $\sigma(\mathcal{E})=\mathcal{E}$, $\sigma \circ r = r \circ \sigma$, and $\sigma \circ i = i \circ \sigma$. The proof of this boils down to the 1-dimensional version of a theorem of Alexander: every homeomorphism of $[0,1]$ that fixes $0$ and $1$ is isotopic to the identity.

To address the question of normality, the image of the natural homomorphism $\mathcal{M}(X) \to [X,X]$ is not a normal subgroup. One way to see this is to use two facts: the group $[X,X]$ acts on the set of conjugacy classes of the free group $\pi_1(X)$; and every conjugacy class is uniquely represented by a "circuit" in $X$, meaning a cyclically ordered sequence of edges without cancellation, modulo cyclic permutation. When $X$ is a finite graph, it contains only finitely many embedded circuits, corresponding to a finite set of conjugacy classes that is invariant under $\mathcal{M}(X)$. But $[X,X]$ contains an element $\Phi$ which takes an embedded circuit to a non-embedded circuit (up to homotopy), hence $$\Phi \mathcal{M}(X) \Phi^{-1} \ne \mathcal{M}(X) $$

In fact if $X$ has rank $\ge 3$ one can show that $[X,X]$ contains an element $\Phi$ that has no periodic conjugacy classes. If $X$ has rank $2$, then the conjugacy class of $aba^{-1}b^{-1}$ is periodic for every element of $[X,X]$, and similarly for $ab^{-1}a^{-1}b$, and for the inverses of those two, and for all their powers. However, all you have to do is to take one embedded circuit in $X$ and find one $\Phi \in [X,X]$ which takes that embedded circuit to a nonembedded one, and that's pretty straightforward.

By the way, I like this normality question, I'm going to make it an exercise in my $\text{Out}(F_n)$ book.

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  • $\begingroup$ Thank you! I’m glad that you liked the question. Two questions: 1. What is an embedded circuit as opposed to one not embedded? Also, is rank just the the number of generators for the free group associated to the graph? $\endgroup$ – Andres Mejia Nov 6 '17 at 14:59
  • $\begingroup$ Yes to the second question. For the first, for example in the $a,b$ rose, $a$ and $b$ are embedded circuits, $ab$ is not embedded. $\endgroup$ – Lee Mosher Nov 6 '17 at 15:05
  • $\begingroup$ ah, so take $\Phi(a,b):=a \mapsto ab, b \to B$ as an element of $\mathcal{M}(X)$, and consider the permutation $\sigma$ that swaps $a,b$. Then $\Phi \sigma \Phi^{-1}(b)=ab$ so the function is not in $\mathcal{M}(X)$. $\endgroup$ – Andres Mejia Nov 6 '17 at 15:10
  • $\begingroup$ Yup! Plus more random characters... $\endgroup$ – Lee Mosher Nov 6 '17 at 15:21

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