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I'm given the following:

We have the functions \begin{align} a&: \mathbb R^2 \rightarrow \mathbb R \tag 1\\ b&: \mathbb R^3 \rightarrow \mathbb R \tag 2 \end{align} and the function composition is $$ c = a \circ b: \mathbb R^4 \rightarrow \mathbb R \tag 3 $$

But this composition doesn't coincide with the definition from Wikipedia:

The functions $f:X \rightarrow Y$ and $g:Y\rightarrow Z$ are composed to yield a function... The resulting compositie function is denoted $g\circ f : X \rightarrow Z$, defined by $(g\circ f)(x) = g(f(x))$.

From the definition the range of $f$ is the same as the domain of $g$. But this doesn't apply to $(1)$ and $(2)$. Can somebody spread some light of what is going on in $(1)-(3)$?

Thanks!

Update:

I saw this example in a physics book.

We have the differential equation $\frac{d r(t)}{dt}= u(r(t))$ where \begin{align} u&: \mathbb R^3 \rightarrow \mathbb R^3 \tag 4\\ r&: \mathbb R \rightarrow \mathbb R^3 \tag 5 \\ \frac{d r(t)}{dt}&:\mathbb R \rightarrow \mathbb R^3 \tag 6 \end{align}

I guess $\frac{d r(t)}{dt}$ is the composition but it is not correct according to the definition.

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  • $\begingroup$ As laid out, $c$ doesn't make sense. "Composition" means the output of one function feeds the input of another function. So if it were to make sense, the function $a\circ b$ would take an element $x$ of the domain of the function $b$, and feed the output $b(x)$ into the function $a$ (so $b(x)$ must implicitly lie in the domain of $a$ if it is to serve as input for $a$), producing the output $a(b(x))$. This is why one says that $(a\circ b)(x) = a(b(x))$. In your case, $b(x)$ is a real number, not a point in the domain of $\mathbb R^2$ of $a$. $\endgroup$ Commented May 11, 2020 at 15:06

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$a \circ b$ does not exist. $a \circ b$ means the function defined by $(a \circ b)(x) = a(b(x))$. But your $a$ needs to act on a member of $\mathbb R^2$, while $b(x)$ is only in $\mathbb R$.

EDIT: For your physics example, the only composition I see is $u(r(t))$, which could be written as $(u \circ r)(t)$. This is OK because $r: \mathbb R \to \mathbb R^3$ and $u: \mathbb R^3 \to \mathbb R^3$.

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  • $\begingroup$ Thanks for the answer. I updated my question with a second example. Does it make any sense? $\endgroup$ Commented May 11, 2020 at 15:20
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Your second example looks correct to me:

  • $r$ takes a real and outputs a triple of reals.

  • $u$ takes a triple of reals and outputs a triple of reals.

  • So "$u\circ r$" is well-defined: it's "do $r$ first and then $u$," and takes in a real and outputs a triple of reals.

Note that $r$ and $u(r)$ have the same "type:" they're both functions from $\mathbb{R}$ to $\mathbb{R}^3$. In general, the type of a function will be the same as the type of the derivative of that function (assuming the latter actually makes sense!).

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In your first example (with $a$ and $b$) the composition $a \circ b$ is not clearly defined, although it might be that $b$ maps onto the $x$-axis so that the composition is meant to be $(a \circ b)(x) := a(b(x), 0)$ which is a mapping $\mathbb R^3 \to \mathbb R$ if $b : \mathbb R^3 \to \mathbb R$ and $a : \mathbb R^2 \to \mathbb R.$ From where you got $\mathbb R^4$ I cannot understand.

The example under "Update" is however correct: $r : \mathbb R \to \mathbb R^3$ describes the path of some particle, i.e. $r(t)$ says where the particle is located as a specific time $t$, $u : \mathbb R^3 \to \mathbb R^3$ is a vector field, and $u \circ r : \mathbb R \to \mathbb R^3$ expresses how the vector field experienced by the particle depends on time. The equation $\frac{dr(t)}{dt} = (u \circ r)(t) = u(r(t))$ is a differential equation describing a particle that at every moment moves in the direction of the vector field where the particle is located, and with a speed given by the magnitude of the vector field. The solution is the pathline of the particle.

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