Is the languague L={, M accepts a finite amount of words} decdidable? Is $L=\{&lt;M&gt; | L(M) \ is \ finite\} $ decidable ? M is a TM.
I think its relative simple to proof with the theorem of rice. But I am interested in a solution which does not use the Rice theorem.
This my try  :
Let f(&lt;m,w&gt;) be a function which works in the following way :

Run w on M
If M accepts Construct a TM Mwhich accepts only the word w and return M
If M rejects Construct a TM Mwhich accepts everything. Return M

So if m is in $A_{TM}= \{&lt;M,w&gt;|M \ accepts \ w\}$ we know that f(&lt;m,w&gt;) is in L. If m is not in A then we know that f(&lt;m,w&gt;) does accept every word and therefore infinity words. So f(&lt;m,w&gt;) not in L.
Is this a correct mapping reduction ?
The function you defined is not a reduction at all - it may not even stop!
The problem is running $m$ on $w$: can you be sure $m$ wont be stuck in an infinite loop on $w$? you can't.
You can define a proper reduction as follows: (on input $&lt;m,w&gt;$)
Create the machine $M_{m,w}$that does the following algorithm, and return in: (on input $s$)

Emulate $m$ on $w$ for $|s|$ steps. If $m$ halted in that time, reject $s$. Otherwise, accept $s$.

I will leave it for you to prove this is a proper reduction from $H_{TM}$ to $L$ (its a good exercise!)