# Mapping reduction from $A_{TM}$ to $INFINITE_{TM}$ same as to $ALL_{TM}$?

I was trying to solve a problem with a mapping reduction from $$A_{TM}$$ to $$INFINITE_{TM}$$, and came across a solution that was 100% identical to another solution I saw for $$A_{TM} \leq_M ALL_{TM}$$. This is the reduction:

$$M_f$$: Given an input of $$\langle M, w\rangle$$, where $$M$$ is a $$TM$$ and $$w$$ is a word:

Define a TM $$M_1$$:

Given the input $$x$$ do:

1. Run $$M$$ on $$w$$ and return its result
2. Return $$\langle M_1\rangle$$

This is correct for $$ALL_{TM}$$ because if $$M$$ accepts $$w$$, $$L(M_1)=\Sigma^*$$, whereas if it rejects, $$L(M_1)=\emptyset$$

However, it is also given as a solution for $$A_{TM} \leq_M INFINITE_{TM}$$, where

$$INFINITE_{TM}=\{\langle {M \rangle} |$$ $$M$$ is a $$TM$$, $$L(M)$$ is an infinite language $$\}$$, for example here: http://www.sfu.ca/~kabanets/308/lectures/lec7.pdf

I don't fully understand why theirs, or other similar explanations, are correct. What about infinite languages that aren't $$\Sigma^*$$? What about finite languages that aren't $$\emptyset$$?

Language $$A$$ is m-reducible to $$B$$ (denoted $$A\leq_mB$$) if there is a computable function $$f:\Sigma^* \rightarrow \Sigma^*$$ such that for every $$w\in \Sigma^*$$, $$w\in A \iff f(w)\in B$$
The point you should pay attention to is that there is no need to talk about every $$x\in B$$. If you can map all elements of $$A$$ to one special element of $$B$$ say $$x_0$$ and all strings that don't belong to $$A$$ to string which doesn't belong to $$B$$ say $$y_0$$ then you are done!
Maybe if you think of $$A\leq_m B$$ as language $$B$$ is harder than $$A$$ then you feel better about the reduction. This means if you can decide $$B$$ then you also can also decide $$A$$.
So by the proof, you have provided we map every $$\langle M,w\rangle \in A_{TM}$$ to $$\Sigma^*$$ and every $$w' \notin A_{TM}$$ to $$\emptyset$$. Thus $$INFINITE_{TM}$$ is harder than $$A_{TM}$$. If you could decide just one elements of $$INIFINITE_{TM}$$ which is $$\langle M \rangle$$ that $$L(M)=\Sigma^*$$ then you could decide $$A_{TM}$$.