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Let us work in Top, the category of topological spaces - although the reader is welcome to replace this by their favorite convenient category of topological spaces. If $X,Y$ are spaces, let $X\ast Y$ denote the join of spaces, $X\coprod X\times I\times Y\coprod Y/\sim$, where $\sim$ is the equivalence relation generated by $x\sim (x, 0,y)$ and $(x,1,y)\sim y$ for any $x,y$.

A good way to think of a point in the join is in the style of barycentric coordinates - if $X, Y$ are both nonempty, the join is homeomorphic to the set of formal linear combinations $t_0x+t_1y$, with $t_0,t_1\geq 0$ and $t_0+t_1= 1$.

It is not hard to see that $\ast$ is a monoidal bifunctor, with unit $\emptyset$. For the associativity axiom, note that $X\ast Y\ast Z$ (again assumed nonempty)can be naturally identified with the set of formal linear combinations $t_0x+t_1 y + t_2z$, $(t_0,t_1,t_2)$ a coordinate in the standard $2$-simplex; if $t_0'(t_0x + t_1y)+ t_1'z$ is a point in $(X\ast Y)\ast Z$, we send it to $t_0t_0'x + t_0't_1y + t_1'z$.

The singleton set $1$, i.e. the $0$-simplex, is a monoid with respect to this monoidal product. The unit maps $\emptyset\to 1$ and $1\ast 1\to 1$ need not be described, as they are unique. Associativity of multiplication is trivial, etc.

The $n$-fold join $1\ast\dots\ast 1$ can be naturally identified with the $(n-1)$-simplex. By the universal property of the augmented simplex category $\Delta_a$, the category of finite ordinals and nondecreasing maps between them, associated to the choice of monoidal product $\ast$ and and monoid $1$ there is a functor $\Delta_a\to \mathbf{Top}$ sending $[0]$ to $1$, sending the unit $\emptyset=[-1]$ to the unit $\emptyset$, and respecting the monoidal product, up to isomorphism. This functor is the standard embedding of the simplex category $\Delta_a$ into $\mathbf{Top}$ which sends $[n]$ to $\Delta^n$.

Because $1$ is a monoid, the functor $1\ast - : \mathbf{Top}\to \mathbf{Top}$ - commonly known as the cone functor - is a monad in a natural way. The multiplication can be described as follows: If $X$ is a given space, let $p_0, p_1$ denote two singleton sets; then the elements of $p_0\ast(p_1\ast X)$ are represented by formal linear expressions $t_0p_0 + t_1(t_0'p_1+t_1'x)$, where $t_0,t_1,t_0',t_1'$ are all $\geq 0$, and $t_0+t_1=1,t_0'+t_1'=1$. The multiplication identifies $p_0$ with $p_1$ and anything on the line between them, so $\mu : p_0 \ast (p_1\ast X) \to p_2 \ast X$ sends $t_0p_0 + t_1(t_0'p_1+t_1'x)$ to $(t_0 + t_1t_0')p_2 + t_1t_1'x$. If $h : 1\ast X\to X$ is an action, and representing $t_0p_0 + t_1x$ by the pair $(t_0,x)$ for concision, we can write that the associativity axiom of the monad asks that $h(t_0,h(t_0',x))= h(t_0+t_1t_0',x)$

My question is - what are the algebras of this monad? It is clear that an algebra of the monad is a contractible topological space, as the unit map $\eta : X\to C(X)$ sends the point $x$ to $0\cdot p_0 + 1\cdot x$. A left inverse to this map would necessarily, it is clear, contain the data of a contracting homotopy, as it would send the singleton $p_0$ to some distinguished point $x_0\in X$, and to every other $x$ it would associate a path from $x_0$ to $x$.

The associativity axiom is interesting, though. Translated from diagrammatic language into equational, if $h$ is an algebra map, viewed as a homotopy $h:I\times X\to X$, it should satisfy (for some distinguished choice of basepoint $x_0\in X$) $$h(1,x) = x_0$$ $$h(0,x) = x$$ $$h(t,h(t',x)) = h(t+(1-t)t',x)$$ for all $t,t'$ in $[0,1]$.

Now, if $X$ is some convex polyhedron which is a subset of $\mathbb{R}^n$, such a contracting homotopy is easily found: set $h(t,x) = tx_0 + (1-t)x$. So motivated by this example, let us call such a contracting homotopy a "straight-line homotopy." So algebras of the cone monad are precisely the spaces admitting a contracting straight-line homotopy.

My question is: What spaces are these? Is there a nice characterization? Does every contractible space admit a contraction by a straight-line homotopy? (I have tried to show this directly by coming up with some clever elementary argument involving reparametrizing the rate of an arbitrary homotopy, in the style of standard lemmas in Chapter 0 of Hatcher, but so far it evades me.)

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    $\begingroup$ Could you write down an explicit multiplication $\mu$ for the cone monad? I think there's some choice there (up to homotopy?). In the last equation you're getting the $t + (1 - t) t'$ from a particular choice of $\mu$. I wonder if a different (but equivalent) $\mu$ would give you something more easily manageable (something involving $\max$ and $\min$). $\endgroup$ Apr 1, 2020 at 9:23
  • $\begingroup$ Hi Prof. Bauer; let $p_0$ and $p_1$ denote two singleton points. Under the identification of $p_0\ast (p_1\ast X)\cong p_0\ast p_1 \ast X$ I mention above, a point in ${p_0}\ast (p_1 \ast X)$ is a formal linear combination $t_0p_0 + t_1p_1 + t_2 x$ for some $x$, where $(t_0,t_1,t_2)$ is a point in the standard $2$-simplex with its usual embedding in $\mathbb{R}^3$. The multiplication should send this to $(t_0+t_1)p_0 + t_2x$ in $p_0\ast X$. Here $t_0$ and $t_1$ are my $t, t'$ in the original post respectively. $\endgroup$ Apr 1, 2020 at 9:49
  • $\begingroup$ I do not quite understand your comment about replacing $\mu$ with an equivalent one. If we work in Top, $\mu$ should be pinned down up to various commuting homeomorphisms; I had not considered the problem in the homotopy category, perhaps I just assumed that the problem would become trivial there. $\endgroup$ Apr 1, 2020 at 9:50
  • $\begingroup$ I have edited the post to explain how I came up with the multiplication of the monad. $\endgroup$ Apr 1, 2020 at 10:09
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    $\begingroup$ Things will work better if you reverse the direction of $I$ (i.e. conjugate by $t\mapsto 1-t$) and then introduce the notation $t*x$ for $h(t,x)$. Then you have $1*x=x$ and $s*(t*x)=(st)*x$ and $0*x=x_0$, so we just have based spaces with continuous action of the monoid $I$ with the extra condition $0*x=\text{basepoint}$. $\endgroup$ Apr 1, 2020 at 10:18

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