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Just out of curiosity that came from a topology homework assignment where I had to show the composition of 3 injective functions was injective.

Suppose $f_i : A_i \mapsto A_{i+1}$ were injective where $i \in \mathbb{N}$. I know that the composition of n such functions, i.e. $\bigcirc_{i=1}^n f_i$, is injective. But what about $\bigcirc_{i=1}^\infty f_i$? Is this an injective function?

Also, is composition bounded to countability? That is, I think that the definition of composition limits us to an ordering which means that we can't have a composition of uncountably many functions. Is this true?

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  • $\begingroup$ How do you define $\bigcirc_{i=1}^\infty f_i$? $\endgroup$
    – joriki
    Commented Sep 10, 2012 at 16:12
  • $\begingroup$ $\left( \cdots \circ f_n \circ f_{n-1} \circ \cdots \circ f_1 \right)$ $\endgroup$
    – torrho
    Commented Sep 10, 2012 at 16:17
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    $\begingroup$ Yes, I thought as much, but what does that mean? If $f_i:\mathbb R\to\mathbb R,x\to-x$, then what is $\bigcirc_{i=1}^\infty f_i$? $\endgroup$
    – joriki
    Commented Sep 10, 2012 at 16:19
  • $\begingroup$ oh wow. Thats almost analogous to the "chicken or the egg" delimma... So then, perhaps it is not possible to define deterministic infinite composition, correct? $\endgroup$
    – torrho
    Commented Sep 10, 2012 at 16:21
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    $\begingroup$ There is a useful notion of transfinite composition for an arbitrary number of functions. However, the construction is somewhat abstract (and sorry, I couldn't find a more palatable link). If you have functions between sets or topological spaces it turns out that the transfinite composition of injective functions still is injective. $\endgroup$
    – t.b.
    Commented Sep 10, 2012 at 17:47

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Though I'm very late in noticing this, I thought it still deserved an answer.

It's entirely possible to define a "deterministic" infinite composition.

Any sequence of functions has a sequence of left compositions (which was how you were composing them) and a sequence of right compositions. Any topology on the function space determines which sequences of functions converge and which don't. If the topology is Hausdorff, sequences have at most one limit, so you get a workable definition of infinite composition.

(This is the same for any binary operation on a topological space, whether that's addition, composition, or whatever.)

Obviously the function joriki mentioned, $x \mapsto -x$, won't have a convergent infinite self-composition sequence under most topologies you'd care to think about.

The infinite composition of injective functions need not be injective. For instance, the function $f:x \mapsto x/2$ has the self-composition sequence $f_k: x \mapsto x/2^k$ which converges pointwise to the constant zero function.

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    $\begingroup$ Wow! You've resurrected this from the grave and gave a solid answer. Thanks. $\endgroup$
    – torrho
    Commented Jan 16, 2014 at 23:04
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    $\begingroup$ +1 in appreciation for resurrecting a question which I was wondering about myself, and came across post via searching. $\endgroup$
    – user105475
    Commented Jun 6, 2014 at 0:31

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