Let a sequence of maps between topological spaces $$ X_1\to^{f_1}X_2\to^{f_2}X_3\to^{f_3}\cdots $$ The mapping telescope is denoted by $T$. Under what conditions will $H_*(T)$, the homology of $T$, will be the same as the direct limit of the following sequence $$ H_*(X_1)\to^{f_1*}H_*(X_2)\to^{f_2*}H_*(X_3)\to^{f_3*}\cdots? $$ Note that in particular, if $f_1,f_2,......$ are injective, then $T=\bigcup X_i$ and by Homology of mapping telescope or proposition 3.33 in Hatcher's book, homology and direct limit commute.

Origin of the question:

In the paper Homology fibrations and group completion theorem, McDuff-Segal, the space $M_\infty$ is constructed as follows: Let $M$ be a topological monoid with $\pi_0M=\{0,1,2,3,...\}$, $M=\bigsqcup_{j=0}^\infty M_j$ where $M_j$'s are the path-connected components of $M$ such that $M_j$ is the component corresponding to $j\in\pi_0M$. We choose $m_1\in M_1$ and consider the sequence \begin{eqnarray} M\overset{\cdot m_1} \longrightarrow M\overset{\cdot m_1} \longrightarrow M\overset{\cdot m_1} \longrightarrow \cdots \end{eqnarray} From this sequence we can form a mapping telescope $$ M_\infty=(\bigsqcup_{i=1}^\infty [i-1,i]\times M)/\sim $$ where $\sim$ is generated by the relations $ (i,x)\sim (i, x m_1) $ for any $x\in M$ and $i\geq 1$. Assume that $H_*(M)[\pi_0M^{-1}]$ (which is just $H_*(M)[m_1^{-1}]$ in this case) can be constructed by right fractions. Then

why $H_*(M_\infty)$ equals to the direct limit of \begin{eqnarray} H_*(M)\overset{\cdot m_1*} \longrightarrow H_*(M)\overset{\cdot m_1*} \longrightarrow H_*(M)\overset{\cdot m_1*} \longrightarrow \cdots? \end{eqnarray}

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    $\begingroup$ Look at proposition 3.33 in Hatcher's book, for example: if every compact subset of $T$ is contained in some $X_n$, then $\varinjlim H_*(X_n) = H_*(\varinjlim X_n)$. $\endgroup$ – Najib Idrissi Jun 28 '15 at 7:32
  • $\begingroup$ There seem to be two questions here: whether homology commutes with direct limits, and how the homology of the mapping telescope is related to the direct limit of the homology. $\endgroup$ – Zhen Lin Jun 28 '15 at 7:44
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    $\begingroup$ The second question has been answered in the context of topological spaces: Homology of mapping telescope I would hope that it isn't too different for topological monoids… $\endgroup$ – Takumi Murayama Jun 28 '15 at 22:08

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