Verify that if you have an image I with all zeroes except for a single

location with value $1,$ that $(f**I)(a,b)=f.$ Try it for a $33$ filter $f$ and a $55$ image I with a $1$ in the middle and zeros everywhere else. Note that the result $(f**I)(a,b)$ might be a bit offset from $f(a,b)$ depending on your indexing.