A problem regarding $k\circ (f*g)=(k\circ f)*(k\circ g)$.

Question

My Algebraic Topology book states the following:

Let $k:X\to Y$ be continuous path. If $f$ and $g$ are two paths in $X$ with $f(1)=g(0)$, then $$k\circ(f*g)=(k\circ f)*(k\circ g)$$

I'm trying to construct an example to understand this concept better, but going wrong somewhere.

Let $f:[0,1]\to(t,3t)$ and $g:[0,1]\to (t,3-3t)$. Clearly $f(1)=g(0)$. Let $k:\Bbb{R^2}\to\Bbb{R^2}$ be defined as $k:(x,y)\to(x^2,y^2)$. Then the mapping of $f$ will be $f':[0,1]\to(t',9t')$, and $g':[0,1]\to(t',9t'-18\sqrt{t'}+9)$. This can easily be verified. Now $f'*g'=(k\circ f)*(k\circ g)$ is $\left\{ \begin{array}{l l} (t',18t') & \quad \text{for $t'\in[0,\frac{1}{2}]$}\\ (t',18t'-18\sqrt{2t'-1}) & \quad \text{for $t'\in[\frac{1}{2},1]$} \end{array} \right.$

Now we shall determine $k\circ(f*g)$. Here $f*g$ is $\left\{ \begin{array}{l l} (t,6t) & \quad \text{for $t\in[0,\frac{1}{2}]$}\\ (t,6-6t) & \quad \text{for $t\in[\frac{1}{2},1]$} \end{array} \right.$

Hence, $k\circ(f*g)$ is $\left\{ \begin{array}{l l} (t',36t') & \quad \text{for $t'\in[0,\frac{1}{2}]$}\\ (t',36t'-72\sqrt{t'}+36) & \quad \text{for $t'\in[\frac{1}{2},1]$} \end{array} \right.$

So I'm getting $$(k\circ f)*(k\circ g)\neq k\circ (f*g)$$

Could someone please help me out? I realise this might entail a lot of seemingly mundane calculations, but I've been stuck on this for some time now, and would greatly appreciate any inputs.

Answer 1

I assume that $k$ is merely continuous, not a continuous path (that would require $X$ to be an interval). In fact your example $k$ is a continuous map $\mathbb R^2\to\mathbb R^2$, not a path. But your $f$ and $g$ fail to be paths in $\mathbb R^2$. You should have something like $f\colon[0,1]\to\mathbb R^2$, $t\mapsto (t,3t)$ and $g\colon[0,1]\to\mathbb R^2$, $t\mapsto (t+1,3-3t)$

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After your edits, note that $f(t)=(t,3t)$, $g(t)=(t,3-3t)$ and $k(x,y)=(x^2,y^2)$ implies $(k\circ f)(t)=(t^2,9t^2)$, $(k \circ g)(t)=(t^2,9-18t+9t^2)$, hence (of course) $(k\circ f)(1)=(1,9)=(k\circ g)(0)$. No square root gets introduced anywhere.

Answer 2

Firstly, I'm assuming that $k : X\rightarrow Y$ is a continuous map.

In this case, all your identity is saying, is that if you first concatenate your paths (ie, form $f*g$, then use $k$ to send the resulting path to $Y$), it's the same as first sending the paths $f$ and $g$ to $Y$ via $k$ (getting $(k\circ f), (k\circ g))$, and then concatenating the result.

A nice way to think about this is taking $X = \mathbb{R}^3$, $Y = \mathbb{R}^2$, and $k$ to be some projection map. In your head, you should think of $\mathbb{R}^3$ as lying above $\mathbb{R}^2$, and the projection map is basically the ``shadow'' of points/paths in $\mathbb{R}^3$, under some bright light that lies far above you, which shines through $\mathbb{R}^3$ onto $\mathbb{R}^2$.

In this case, drawing the concatenation $f*g$ in $\mathbb{R}^3$, then looking at the shadow, is clearly the same as drawing each path in $\mathbb{R}^3$ separately, looking at their shadows, and noticing that their shadows are joined head to tail!

Answer 3

There are a lot of things wrong. The biggest one is that you are reparameterizing $t'=t^2$. So in your concatenated path $f*g$, $t$ traces from $0$ to $1/2$ in the first term so $t'$ traces from $0$ to $1/4$.