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Show that $f\colon X \rightarrow Y$ is a homotopy equivalence if and only if there exist maps $k,h\colon Y\rightarrow X$ such that $f\circ k$ is a homotopy equivalence of $Y$ to itself, and $h\circ f$ is a homotopy equivalence of $X$ to itself.

First suppose that $f\colon X\rightarrow Y$ is a homotopy equivalence, then there exists a map $k\colon Y\rightarrow X$ such that $k\circ f \sim 1_{X}$ and $f\circ k\sim 1_{Y}$.

Notice that $f\circ k\circ 1_{Y}=f\circ k\sim 1_{Y}$ and $1_{Y}\circ f\circ k=f\circ k\sim1_{Y}$. Therefore $1_{Y}$ is a homotopy inverse of $f\circ k$ and $f\circ k$ is a homotopy equivalence of $Y$ to itself.

Similarly suppose $h\colon Y\rightarrow X$ is a homotopy inverse of $f$ then $h\circ f\sim 1_{X}$ and $f\circ h\sim 1_{Y}$.

Notice that $h\circ f\circ 1_{X}=h\circ f\sim1_{X}$ and $1_{X}\circ h\circ f=h\circ f\sim1_{X}$. Therefore $1_{X}$ is a homotopy inverse of $h\circ f$ and $h\circ f$ is a homotopy equivalence of $X$ to itself.

Now suppose there exist maps $h,k\colon Y\rightarrow X$ such that $f\circ k$ is a homotopy equivalence of Y to itself and $h\circ f$ is a homotopy of $X$ to itself.

So there exists a map $y\colon Y\rightarrow Y$ such that $f\circ k\circ y\sim 1_{Y}$ and $y\circ f\circ k\sim1_{Y}$.

Similarly there exists a map $x\colon X\rightarrow X$ such that $h\circ f\circ x \sim 1_{X}$ and $x\circ h\circ f\sim 1_{X}$.

I was thinking if I can show that $k\circ y\sim x\circ h$ (since homotopy inverses are unique up to homotopy) then we have shown that there exists a homotopy inverse and therefore $f$ is a homotopy equivalence. But I cannot find a homotopy from $k\circ y$ to $x\circ h$. I have tried to find one using the homotopies from $f\circ k\circ y$ to $ 1_{Y}$ and from $x\circ h\circ f$ to $ 1_{X}$ but didn't have any success.

Any advice would be appreciated, thank you.

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It's probably easiest to introduce the homotopy category, so as to avoid having to keep track of individual homotopies. –  Zhen Lin Apr 12 at 21:40

1 Answer 1

I have not covered homotopy category yet so cannot use it I'm afraid. However I have found the solution I think...

Consider the following $$x\circ h\circ f \circ k\circ y$$ We know that $x\circ h\circ f\sim 1_{X}$ so if we substitute this into the above we get that $$x\circ h\circ f \circ k\circ y\sim 1_{X}\circ k\circ y\sim k\circ y$$ Similarly we know that $f\circ k\circ y\sim 1_{Y}$ so we can substsitute $1_{Y}$ to get the following $$x\circ h\circ f \circ k\circ y\sim x\circ h\circ 1_{Y}\sim x\circ h$$ Therefore $x\circ h\sim x\circ h\circ f \circ k\circ y\sim k\circ y$, hence $x\circ h\sim k\circ y$ as required.

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