# Specifying a homomorphism by given the images of the generators and extending "lineary"

Let $M$ be a finitely generated monoid with generators $g_1, \ldots, g_k$. Now is every homomorphism $\varphi : M \to N$ uniquely specified by listing the images of its generators? Of course the image of every element is given by extension, but what if some element does not has a unique decomposition? Suppose we have $m = g_1 g_2 = g_3 g_4$, and an image $\varphi(g_i)$ for each $i = 1,\ldots, k$ is given. Then what if $\varphi(g_1)\varphi(g_2) \ne \varphi(g_3)\varphi(g_4)$? Then $\varphi$ is not a homomorphism?

• Indeed, the homomorphism is uniquely determined if it exists. Apr 19, 2016 at 18:49
• Yes, even before $\phi$ is a homomorphism, it must make sense. This is what's called checking for well definedness. Apr 19, 2016 at 19:01

Thanks for the comments. Just an example which shows that giving the images of generators is not enough to give a homomorphism. Consider the groups $G = \{1,a,b,c\}$ with $a^2 = b^2 = c^2 = 1$ and $ab = c$, and $H = \{1,g,g^2,g^3\}$, then $G$ is generated by $a$ and $b$, but the map $\varphi(a) = g$ and $\varphi(b) = g$ is not a homomorphism, as for example $1 = \varphi(1) = \varphi(aa) = g^2$, a contradiction.