Skip to main content
added Akhil's proof of the missing pushout
Source Link

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array} $$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too (I'm blanking on how to do that at the moment), we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.

Here's Akhil's proof that the left hand square is a pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{\pmatrix{0&1}} & X \oplus X & \xrightarrow{\pmatrix{0 & 1 \\ 1 & -1}} & X \oplus X \\ \downarrow & & \downarrow \rlap{\scriptstyle\pmatrix{1\\ 0}}& & \downarrow \rlap{\scriptstyle\pmatrix{1\\1}}\\ 0 & \to & X & \xrightarrow{1} & X \\ \end{array} $$

The outer rectangle is what we'd like to show is a pushout. The right hand square is a pushout because both horizontal arrows are isomorphisms, and the left hand square is a pushout because it is the (pointwise) direct sum of these two (where all the maps $X \to X$ are the identity):

$$ \begin{array}{cccccc} 0 & \to & X & & X & \to & X \\ \downarrow & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & & 0 & \to & 0 \\ \end{array} $$

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array} $$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too (I'm blanking on how to do that at the moment), we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array} $$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too, we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.

Here's Akhil's proof that the left hand square is a pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{\pmatrix{0&1}} & X \oplus X & \xrightarrow{\pmatrix{0 & 1 \\ 1 & -1}} & X \oplus X \\ \downarrow & & \downarrow \rlap{\scriptstyle\pmatrix{1\\ 0}}& & \downarrow \rlap{\scriptstyle\pmatrix{1\\1}}\\ 0 & \to & X & \xrightarrow{1} & X \\ \end{array} $$

The outer rectangle is what we'd like to show is a pushout. The right hand square is a pushout because both horizontal arrows are isomorphisms, and the left hand square is a pushout because it is the (pointwise) direct sum of these two (where all the maps $X \to X$ are the identity):

$$ \begin{array}{cccccc} 0 & \to & X & & X & \to & X \\ \downarrow & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & & 0 & \to & 0 \\ \end{array} $$

added confesion about left square
Source Link

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array} $$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too (I'm blanking on how to do that at the moment), we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array} $$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too, we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array} $$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too (I'm blanking on how to do that at the moment), we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.

Source Link

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$ \begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array} $$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too, we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.